 So, in today's session of binomial theorem, we are going to primarily focus on the binomial coefficient and the binomial coefficient series, okay. So, we are going to focus on binomial coefficient and binomial coefficient series. So, I will be starting with binomial coefficient properties. I will say sir, we have already done this in our permutation combination. I agree you have already done this, but it's very important for us to know them again or revisit them again at least in order to make a progress in this chapter, okay. So, what are the properties that you have learned with respect to binomial coefficient? The first property that you already are aware of NCR and NCN, sorry NC0 and NCN both have the same values, okay, very simple property. Next property is, next property is NCR and NCN minus R are same values. This is also known to you the number of ways to select R distinct objects from N distinct objects is the same way as the number of ways to select N minus R objects which you actually don't want to select, okay. Writing NCR in terms of the lower coefficients, okay. So, you can write NCR as N by R times N minus 1 CR minus 1. You can further write it as NN minus 1 RR minus 1 times N minus 2 CR minus 2 and this can continue on and on. This can continue on and on, okay. N minus 3 CR minus R minus 3, okay. I'm just writing dot, dot, dot just to show that it can continue on and on. Next, NCR upon NCR minus 1 is N minus R plus 1 by R, very, very important property. Especially when you need the ratio of these binomial coefficients, this property is very, very useful, okay. Pascal's triangle, okay. So, everybody knows Pascal's triangle. NCR plus NCR minus 1 is equal to N plus 1 CR. This is your Pascal's triangle. So, guys and girls, we have already done this. This is called the Pascal's identity. So, we have already done this. We are not, you know, doing it again. We are just quickly recapping it so that if at all we need it, we can use it, okay. Next, RCR, R plus 1 CR, R plus 2 CR, all the way till NCR, okay. Okay, anybody who knows what is going to be the result of this? Anybody who knows what is going to be the result of this? Have you seen this in a series ever before? I think Harshita should be able to answer this. Ah, R. R or R plus 1. So, it's N plus 1, see R plus 1, okay. Now, this is actually called the hockey stick identity, okay. This is what we call as the hockey stick identity. Hockey stick identity. Now, how it actually works? How it actually works? If you add these two, now see here, this RCR, you can actually write it as R plus 1 CR plus 1 because that's actually 1, right. If you add these two, okay, you'll end up getting, you'll end up getting R plus 2 CR plus 1, correct. Now, if you add these two, that'll give you R plus 3 CR plus 1. And this will continue till you reach, till you reach, the term before this will be NCR plus 1, correct. So, if you add these two, you'll get N plus 1 CR plus 1, is it fine? So, this is the small proof for this. I'll be just, you know, removing this off just to make it look a little bit more nicer, but this is the proof for it. So, I'll be rewriting it again, not to worry. Okay, so this will give you R plus 2 CR plus 1, then R plus 2 CR and R plus 2 CR plus 1 will actually give you R plus 3 CR plus 1, okay. And this will continue till I reach, till I reach NCR plus 1 and NCR, which is your last term, that will be N plus 1 CR plus 1, which is your right hand side. Okay, so just keep it in your mind in case you need it any time to solve a question. Now, coming to some important series that we are going to come across in our discussion in today's session. So, I hope everybody has noted these first six results. Okay, done. Okay, so these are the series which everybody should be aware of. I mean, I've already discussed this in our permutation combination chapter, but I'm revisiting it again. NC0, NC1, NC2, all the way till NCN is actually 2 to the power N. Now, how does this result actually come out? Very simple. So, in this question, what do you do? You just write 1 plus X to the power of N. Okay, 1 plus X to the power N, if you write, this is going to be your expansion. Okay, just put X as 1 in both the sides. So, when you put in both the sides, you will end up getting NC0, NC1, NC2, etc. till NCN. Okay, so this is nothing but 2 to the power on this side and you end up getting NC0, NC1, etc. till NCN. So, this result is going to be very, very important. At every look and corner, you will find that this result is going to be utilized in solving many questions. Okay, so please make a note of this, everybody. Okay, now based on the same approach, we have another result which is again known to you, but still I'm rewriting it for you, NC0, NC2, NC4 till wherever it allows it to go. This will be same as NC1, NC3, NC5, etc. Okay, how does this result actually come about? And of course, these two results will be equal to 2 to the power of N-1. So, what is the proof for this result? What is the proof for this result? Let's look into it very, very quickly. We are already aware of it. So, in this result, what I'm going to do? So, this is the result which everybody knows. Let me write one more. So, in this result, we are going to put our X value as a minus 1. Okay, so when you do that, the left-hand side becomes a 0. Right-hand side will start getting these terms with alternative sign. Okay, now what is going to happen? Just send the odd positions on one side, I'm sorry, odd positions on one side like this and even positions on one side. Okay, so this proves the first half of the proof. Now, what is the value of each one of them? Very simple. See, let's say I call this as X. Okay, let's say I call this as X as well. Okay, from property number 7, we see that when we add everything, we get 2 to the power n. That means from property number 7, I can say X plus X is nothing but 2 to the power n, which means 2X is 2 to the power n. So, X is 2 to the power n minus 1. Right, so each one of them is actually equal to 2 to the power n minus 1. Is it fine? Any questions? Any concerns? Any questions? Any concerns? So, actually, I was checking your notes of binomial theorem in the last session. We had completed till divisibility, right? If I'm not mistaken, we had just completed till divisibility questions, correct? Okay, so there are a few more concepts that we have to take up before we start using binomial theorem, before we start talking about binomial theorem series. I thought we had completed those miscellaneous things. So before I start with binomial coefficient series, and before we start using all these eight properties that we have discussed, we have some concepts left off that we need to finish it off. Okay, so let's note this down and let's start with those concepts. Which we have to finish it off, okay? All right, so let's begin with some questions. So we had completed, yeah. Okay, so these are some miscellaneous questions that we need to touch upon. Okay, let's take this question. I hope this is legible to everybody. Show that the integral part, show that the integral part of this expansion is odd. Okay, where n is a natural number. That means if you put any natural number power on this expression and you expand it, you are going to not expand it, you just calculate the value of it. Let's say if you put a value of n as a 1, okay, you'll get a number. That number's integral part will be an odd number. If you put 2 also, you will get a number. That number's also integral part will be odd number. So like that, any natural number power if you put, you will always get an odd number in the integral part of your answer. That means your answer would be some odd number dot something or decibel some value, okay. How do you prove this kind of a question or how do you show this kind of a question without actually citing examples? Should I give you one minute to try this out? Would you like to try this out? Yes, anybody, any idea how to proceed? Okay, all of you please pay attention. See, let's say this expression, this expression n being a natural number gives you i plus f, f is some fraction part, okay. f is some fraction part which lies between zero and one, okay. And i is some integer, okay. And i is some integer, i is some integer, okay. Actually I've written i that itself means integer, so no need to explain what is i here. Now let us take, let us consider an expression like this, okay. Now I claim that this expression will always be a fraction. Okay, I claim that this expression will always be a fraction and that to a non-zero fraction. Now why? See very simple. 25, root 25 is 5 and root 24 is this expression, correct. So the gap between root 25 and root 24 has to be lesser than one. Please note that in order to have a gap of one and more, the difference should be of root 25 and root 16. That means 5 and 4. Then only the gap will become equal to one. So this gap is actually a fraction. And if you raise that gap to a natural number power, that's still be a fraction because fraction raised to a natural number is again a fraction, okay. Now let us do one thing. Let us add the two, okay. Let us add the two. So when you add the left hand side and the right hand side, again I will not write everything down. Please note that since the sign in between is opposite, every second term will start getting cancelled. That means 5 to the power n will come in both the expressions. But the next term which is nc1, 5 to the power n minus 1, 2, root 6. That will get cancelled off, okay. So the term which is going to be coming immediately after 5 to the power n will be nc2, 5 to the power n minus 2, 2 root 6 square, okay. Again, the next term will be nc4, 5 to the power n minus 4, 2 root 6 to the power of 4 and so on and so forth, okay. So on the right side you will be left with i plus f plus f dash, am I right? Any questions here? Now one important thing you would note down here that every term involved here will be an integer. Every term involved over here will be an integer. Check it out. Yes or no? Why every term will be an integer? Because see n, c, r will anyways be an integer, right? 5 to the power a natural number will anyways be an integer. But this irrational term which is 2 root 6 is always subjected to even power if you see it. That means there will not be any irrationality involved. So it will always be a rational. In fact, it will always be integer. So all these terms will be integers, integers, integers. In short, you'll end up getting 2 into an integer which is actually an even integer, okay. And on the right side you will get i plus f plus f dash, yes or no? Now, all of you would appreciate this fact that if this is an integer and this is an integer, whether even or odd, this is an integer, right? And this is also an integer. So from here, can I say f plus f dash will also be an integer? Agreed, because only an integer added to another integer can give you an integer, right? Now, let's come to this expression that I have written over here and this expression which I have written over here. If you add the two, you will end up getting f plus f dash to lie between 0 and 2. Add these two. If you add these two from here, you can make this conclusion. Now, the only integer which lies between 0 to 2 has to be 1, right? Because here you are claiming that f plus f dash is an integer and f plus f dash is between 0 to 2. So the only integer which lies between 0 to 2 is a 1, right? Which means an even integer is equal to the integral part plus 1, which means i has to be even integer minus 1, which can only be an odd integer. Yes or no? Is it fine? So this proves that the integral part of 5 plus 2 root 6 to the power n will always be odd. Now, this question is a show that question, but it can always be objectified. So be aware of these kind of questions. Okay, so we'll take one more because this was just a demonstration for you all. So one more question we'll take up on this. So first make a note of this. Anywhere you want me to take the screen, do let me know. Okay, all right. So let's take one more question. Can I move on to the next slide, by the way, if you're all done with this? Okay, let's take this question. Show that the integral part of this expression is even. So this is a question which is slightly framed in a different way, but the end objective of the question is similar to what we had. You have to show that the integral part is an even number. Done? Done. Achyentya is done. Awesome. Good, Erwin. Okay, so in this case, the approach is slightly different. Here, again, if you see, this is root of 125, isn't it? And 11 is like root of 121. So gap between them is lesser than 1, for sure. So let us first write this down as some i plus f, where f is some fraction between 0 to 1. Okay, now let's consider this term, root 5 minus 11 to the power 2n plus 1. And as I told you, this term will be definitely be a fraction. So this will be somewhere between 0 to 1. Okay, now this time what I'm going to do, I'm going to subtract the result. Now, what makes me subtract it? So see, I want to get rid of any kind of an irrationality. An irrationality will come because of this 5 root 5. In the previous one, it came because of the second term. Here it is coming because of the first term. Okay, so in order to remove those terms where the first term is having an odd power, we have to subtract the result. So you have to take this call when you are trying to solve this question. Whether to add it or whether to subtract it, the core principle behind this would be to get rid of any irrational term or any term which has got that irrational term as an odd power on it. Okay, getting my point. All right, so if you remove it, please understand which terms will get doubled up. The first term will get cancelled. The second term will double up. And what is the second term? What is the second term? 2n plus 1 c1 5 root 5 to the power 2n into 11. Okay, then the fourth term will double up, which is 2n plus 1 c3. 5 root 5 to the power of 2n minus 2 into 11 to the power of... Sorry, 11 to the power 3. Okay, I'm not writing all the terms. And on this side, you'll end up getting something like this. Okay, so each one of these terms here are integers. So this is also an integer. This is also an integer and all these terms will be an integer. So the right side will be 2 into an integer. Okay, sorry, right-hand side is this. Left-hand side will be 2 into an integer. Sorry about mixing left and right. So left-hand side is an even integer. Okay, and right-hand side will be i plus f minus f dash. Okay, now, whether even or not even, it is an integer. And this is already an integer. So that means this guy should also be an integer. Now, if you subtract these two, so if you subtract these two from here, I can say f minus f dash should be between minus 1 to 1. Check it out. Why minus 1? Because let's say you have a value which is slightly less than 1 for f dash and slightly greater than 0 or equal to 0 for this guy. So then the difference will be minus 1. And the max difference will be when this guy becomes very close to 1 and this guy becomes very close to 0. So the difference will be actually between minus 1 to 1. So in light of this, can I say f minus f dash can be only be 0? Because the integer, which is between minus 1 to 1, can only be 0. So this guy will be a 0, which means i plus 0 is equal to an even integer. Okay, that means i is even. Is it fine, any questions, any concerns? Is it fine, any questions? Okay, would you like to try one more or is it good enough? Depends upon your understanding. Do you want one more question of this type or is it good enough? You tell me. Why is the left side even? Because 2 into an integer will be an even integer because there will be a doubling of the terms. I don't need to say sorry, it's okay. Okay, you're just clarifying your doubt. Okay, since nobody is asking anything, nobody is saying anything, we'll take one more question. So let's take this one. If 7 plus 4 root 3 to the power n is s plus t, where n, where this n is a positive integer and s is also a positive integer. So they should actually write an i over here, but they have written an s, it's okay. So n and s both are positive integers, means natural numbers and t is a proper fraction. Show that 1 minus t into s plus t is 1. 1 minus t into s plus t is 1. Very simple, very straightforward question. Very similar to the first time that we did today. Yes, excellent. Two of you are done. Okay, great. See again, the approach is very similar. I mean, if you take this as s plus t, okay. And let's take another term, 7 minus 4 root 3 to the power n. By the way, again, please note 4 root 3 is root of 48. 4 root 3 is root of 48, okay. And 7 is root of 49. So gap between them has to be a fraction. It cannot be more than 1. It cannot be equal to 1 also, right. So this will definitely be another fraction. Let's call it as t dash. And now, since the irrationality is involved in the second term, you add them. That's how your irrationality will go off. So that is going to give you terms like this. So this is going to be an even integer, okay. And this is going to be an integer plus this expression. And as I already told you, t lies between 0 to 1 and t dash also lies between 0 to 1. So their sum will lie between 0 to 2. The only integer which lies between 0 to 2 is a 1, okay. That means, everybody please understand here, that means t plus t dash is 1. That means t dash is 1 minus t. So this 1 minus t here you see it's actually your t dash, okay. So the question setter is asking you to prove that t dash, t dash, which is this term, into s plus t, which is this term, is actually a 1, which is true in fact, because 49 minus 4 root 3 square to the power of n is actually 49 minus 48 to the power of n, which is 1 to the power n, which is your right hand side. Getting this point, okay. So this could also be objectified. They can ask you what is 1 minus t into 1 plus t, which of the following option, you can always mark that. All right. Now one type of question which is again left off is your remainder type of question, remainder problems, okay. So how does binomial theorem help us to solve remainder problems? So I'll just take a small, you know, question over here. See, let's say I want to find out, let's say the question is find the remainder, find the remainder, when I'll just take an example, let's say 7 to the power 100 and let's say, let's say 7 to the power 100, okay. Find the remainder when this is divided by, this is divided by, let's say 25, okay. So how do we solve these kind of questions? Let us try to have a look into this. So there's a small process that we normally follow. Okay, by the way, those who have already done linear congruences, how many of you who prepared for a PRMO, when you were in grade 10 or 9, have come across the concept of linear congruency? Anybody who knows linear congruency, the properties of linear congruency? No, nobody. Yes, Tushar sir would have definitely taught you that. So if you are quite familiar with that, you can use it. Okay, but those who have not learned it, for them, I'll be telling you the binomial theorem approach. Linear congruency, I will not be teaching you because again, it will take around half an hour, 45 minutes, which is definitely not required. See here, the process is, we try to express this expression, okay, as a to the power of B. Okay, let it be multiplied by some number. Let's just see, no problem. Where this expression or this base A is the nearest multiple of 25 plus minus 1. Okay, if possible, I'm saying, if writing like this is not possible, then there is another way to solve this. But if possible, try to write your A, which is the base as a nearest multiple of 25 plus minus 1. Okay, now let me demonstrate that here. So in this case, can I write this guy as 7 to the power 2, which is 49 to the power of 50? Correct, where 49 is a nearest multiple, it's a multiple, so as to say, not a nearest one, it's a multiple of 25 minus 1. Okay, so I can write this as 50 minus 1 to the power 50. So what is the advantage of expressing it like this? So see, if you expand it, you'll get 50 to the power 50 minus 50 C1, 50 to the power 49 plus 50 C2 into 50 to the power 48, and the last term will become a plus 1. Correct me if I'm wrong. Okay, just let me write the term before it, so it'll be minus 50 C49 into 50. Okay, now see here, till here, all the terms are divisible by 50. And hence, and hence divisible by 25 also, because anything which is divisible by 50 will also be divisible by 25. So can I not write this as 25 lambda plus 1? In other words, can I say the remainder that you will be getting when you divide that expression by 25 is going to be 1? Is it fine? Any questions? Any questions? Any concerns? Any questions? Any concerns? Clear? All right, so let's take this question. I mean, a rhyming question with this. The question is, find the remainder when 17 to the power of 256 is divided by 10. Find the remainder when 17 to the power 256 is divided by 10. Done? Okay. Satyam, very good. See, by the way, when somebody asks you what is the remainder when you're dividing it by 10, he's actually asking you the last digit. Okay, so most of you would be aware that when you're finding the last digit of such kind of an expression, we divide the power by 4. Okay. And whatever remainder comes from there, other than 0. Okay, if a 0 remainder comes when you divide this guy by a 4, then you use a 4 and you raise that number to or on this last digit. For example, let's say 256. Okay, when you divide 256 by 4, what do you get? 0. Correct? That means raise 7 to the power 4. Whatever is the last digit of that, that is going to be your answer. And 7 to the power 4, the last digit will be, I mean, something 1. Okay, so this will be a remainder. Plain and simple. That's one way to look into it. And let's say if you want to solve it by our binomial approach. So as I told you, write this guy as something which is a multiple of this number plus or minus 1. So the best answer, the best way to write is 289. 289 comes from 17 square. Correct? I'm so sorry. This is 128. Yeah. So 289 is a multiple of 10 minus 1. Correct? And when you expand it, you get 290 to the power 128 minus 128 C1, 290 to the power 127, 128 C2, 290 to the power 126. The last term will be, I mean, the pen ultimate term would be negative 128 C127 times 290. And the last term will be 1. Okay, plus 1. Remember, plus 1. Now, the sign here matters. Okay, if it was a minus 1, things will change. How will change? I'll tell you. Now all these terms here till here, they will be a multiple of 290. Correct? And hence it will be a multiple of 10. That means it will be some 10 alpha plus 1, which means the remainder is going to be a 1 here again. Okay. So many students asked me, sir, what will happen if your last, this guy is not getting formed? What to do? Okay. I will talk about that also. But let me also tell you what will happen if this number was minus 1. If remember, this number was minus 1, your remainder would be, I will tell me. Let's say there was a minus 1, then the remainder would be 9 exactly. So please note that if you are, let's say hypothetically speaking, if you get a number of this nature, then what do you have to do? You have to release a 10 from here. So that number 9 will become your remainder, not minus 1. Never write a remainder as a negative number. Reminder is always from 0 till 1 less than that number. Okay. So it can never be a negative. It can never be equal to the divisor. It can never exceed the divisor, that is for sure. But let's say if this kind of a scenario comes, your remainder is going to be 9. Your remainder is going to be 9. Are you getting my point? Okay. Now I'll be taking a small question for you, which just give me a second. Who is that question? Yeah. 7 to the power 100 when divided by 90. Yes. So let's take this question. Find the remainder, find the remainder, find the remainder. When 7 to the power 100 is divided by, is divided by, I have such a short memory. What was it? Divided by 13. 13. Okay. Yeah. It will be divisor minus 1. That depends upon what you're devising. But in our example, it was 9. That is what I was trying to say. Okay. Yes. So in this case, how do you solve this question? Find the remainder when 7 to the power 10 is divided by 13. Now here, I don't know whether you are trying it or not, but however you try, you will not be able to express 7 to the power 100 as some base, which is a multiple of 13 plus minus 1. You will try it out. Let's say if you take 7 to the power 2. Okay. So it's 49. 49 is not a multiple of 3 plus minus 1. You can try it out. Okay. If you take 7 to the power 3, which happens to be 243, right? 7 to the power 3, 343. That also you will not be able to write. Okay. Check it out. I mean, first convince yourself. So how do we deal in such cases? So what do we do here? Let us write this as 7 square to the power 50. Okay. So this is actually 49 to the power 50. Okay. So 49 to the power 50. If I write it as a multiple of 13. So what is this? 13 to the power 452, right? So can I write it as 52 minus 3 to the power 50? Correct? Okay. So if I write 52 minus 3 to the power 50, you'll end up getting 50 C0, 52 to the power 50. Then you'll get minus 50 C1, 52 to the power 49 into 3. Okay. Then plus 50 C2, 52 to the power of 48 into 3 square. Okay. And till last term is 50 C49, 52 to the power of 1 into 3 to the power of 49. And the last term that will be left off. By the way, there will be a negative sign here. Okay. And the last term that will be left off will be 3 to the power 50. Right? Now the problem here is till here, everything is divisible by 52. And hence, and hence divisible by 13 also. So you have something like 13 lambda plus 3 to the power 50. Right? So here, this term is not a simple term. It's a huge number. So now your problem comes, your problem simplifies to you finding the remainder from 3 to the power 50. Because whatever remainder comes from this term, that will be your remainder of the original question also. Because other terms are not going to give a remainder. So the problem, you know, boils down or simplifies to finding the remainder. What is the remainder when when 3 to the power 50 is divisible by 13? Okay, let's find it out. Now here, the things will be slightly on the easier side. The reason me, you can write 3 to the power 3 as 27, which is 2 into 13 plus 1. Okay, now see what I'm going to do. So I'm going to break this 3 to the power 50 like 3 to the power 2 into 3 to the power 48. 3 to the power 2 is 9. And this 48, I will write it as 3 into 16, something like this. Now this 27 term is nothing but 26 plus 1. Okay, now don't worry if a nine comes here, that is fine. I can manage with that nine. But I just want to get this as a multiple of 13 plus or minus one, which I was successful in doing. So when you expand this, you end up getting 26 to the power 16. Next term will be 16 C1, 26 to the power 15. Then the next term will be 16 C2, 26 to the power 14, the last term is going to be a plus 1. Okay, so overall, you end up getting something like this. It'll go up till this term plus a nine plus a nine. Yes or no? So till here, everything will be divisible by 13. In fact, everything will be divisible by 26. So it will be divisible by 13 as well. Correct? Let's say I call it as alpha and you'll be left with a nine. That means nine will be your remainder. So the answer to that original question also is going to be nine only. So what is the remainder when this is divisible by 13? The answer for this question will also be nine. Got it? Okay. So maybe in one go, you will not get the answer, but you have to, again, take the remainder over here and again work in the same way. And it may take more than two iterations also. So be prepared for it. Is it fine? What was the process adopted to solve this question? Okay. So with this, we are now going to start with, that's the point Satyam. The bigger the, you know, left part is taken, you know, see it depends upon your understanding. Let's say you have taken 10. Okay. The last term would have been 10 to the power something. Now from there, you have to write it as again a nearest multiple of as a multiple of 13 plus minus one. And it is up to you to do that. So whether you write a 10, whether you write a, you know, whatever number there, you have to face the brunt of this last term that you're going to get, that is three to the power 50. So you will end up getting 10 to the power some power. So you should be able to write again that 10 to the power as a multiple of 13 plus minus one bomb. If you're able to do that, go ahead with your approach. So there's nothing right or wrong in that. And if you start with that, it may take multiple iterations. Okay. So that is what you have to take into your consideration. All right. So we are now going to go towards our binomial coefficient series. In fact, we'll take one question on binomial coefficient properties. So the question goes like this, in the expression x plus y to the power n, the second term is given to you as 240. Third term is given to you as 720. And fourth term is given to you as 1080. Okay. Find the value of x, y and n. Okay. And being some whole number. In the expression x plus y to the power n, the second term is 240. Third term is 720. Fourth term is 1080. The term means the value of the term, the whole value. So you have to find the value of x, y and n. Yes. Anybody see what will be t2? t2 is t1 plus one, right? That will be nc1 x to the power n minus one y to the power one, right? So this is given to us as 240. Okay. Similarly, t3 will be t2 plus one, which is nothing but nc2 x to the power n minus two y square. t4 will be t3 plus one, which is nc3 x to the power n minus three y cube. Okay. So these values are given to us. Okay. Let's call them as one, two and three. Right. Let us divide two by one. So when you divide two by one, you get nc2 by nc1 into y by x, which is going to be three. Let's call it as four and you divide three by two. So that will give you nc3 by nc2 into y by x, which is nothing but now 1080 by 720 will be three divided by two. Okay. Let's call it as four. Oh, very good. Harshita has got n value. Great. Okay. We'll check Satyam. Good try. Okay. Now, having got this, please understand here that nc2 by nc1, which formula we are going to use, which property we are going to use, ncr by ncr minus one, which is n minus r plus one by r. Okay. So this times y by x is three. Correct. And this is going to be n minus three plus one by three times y by x. That is going to be three by two. Okay. Take the ratio of these two as well. So that is going to give you n minus one by n minus two. And I think three by two will come here. That is going to be a two. Okay. So what I'm going, I'm just doing four divided by five. Correct me if I'm wrong. Which means three n minus one is four n minus two. Okay. Which clearly means n value is going to be a five. Check it out. Okay. So n value comes out to be a five. Any questions, any concerns? Okay. So once n value comes out to be five, you put it in, put this in fourth equation. Okay. So here if you put n value as five, you'll end up getting five minus one, which is four, four by two times y by x is equal to three. Which means y is equal to three by two x. Correct. Y is equal to three by two x. Now put this in your first equation. Equation number one. So put this in one. So as per one, nc one, which is five c one. Okay. So let me write that again down. nc one x to the power n minus one y to the power one. nc one is five. And this is going to be four. Y is three by two x. That is 240. Which clearly means five c one x to the power five. That will be four eighty by three, which is one sixteen. Correct. And five c one is five. So it'll be 32. 32. That means x value is a two. If x value is a two, y value is a three. So there you go. X value, y value are all now out. So answer is going to be n is five, x is two, y is three. Is it fine? Any questions? Any concerns? Okay. So this is a very, very basic type of question and it comes a lot in your comparative exams also. But this is not a J main type of question. You can count this as a regional interest exam type of question. So with this, we are now going to start with our binomial coefficient series questions. Okay. So let's start binomial coefficient series. I hope everybody had copied the previous solution and nobody had any doubt. So let's start with binomial coefficient series. Okay. And of course, you will not be getting very simple question based on the same. So let me begin with, you know, one of the coefficient, binomial coefficient series question. Let's say we take this question. Okay. It's a proof that question, but it can always be objectified. By the way, many people have this feeling. So if it is always, if it is an objective question, then I can always substitute the value of, you know, n and get my answers, right? In any type of series question, this is the normal feeling everybody gets. Isn't it? But let me tell you, if you are smart, the J or NTA is smarter. Okay. So they know that this problem can be solved by merely substituting the value of n, whatever is, you know, required to sum up and you can always match with the options. So what they normally do in order to avoid people using these approaches, they will say that this expression is some function of n, like that they will give you. And they will say what is the limit of that function as n tends to 5 or as n tends to, you know, infinity, or maybe they will say what is the integration from, you know, with respect to n from this value to this value, anything they can do with that. Okay. So please do not depend upon the approach of you substituting. Of course, if it works, why not? Use it. If it works, I'm not asking anybody to not take that method. But if it doesn't work, then you should be ready with these concepts as well. Okay. Now let me start with this question as a sample question to begin with. See here, they have given that there is a particular series 1 plus x to the power n whose expansion they have written like this. By the way, most of you would be wondering why did not they write nc0 like this, nc0, nc1, nc2, etc. See, ideally, yes, they should have written it. But normally this n, since it is present everywhere and it is understood, it is very obvious that it is going to be n because the power here is n. So they sometimes drop that particular n. They don't write it. In fact, they start writing ncr as cr just to save their time. Okay. And you will find this particular methodology used in many questions also. They will say, let ncr be cr just to save their typing effort. So now we have to sum up this particular series. Let's say I'm not aware of my answer. Okay. I'm thinking that this answer is hidden from me. So what I need to find out, what is nc0, 2 into nc, sorry, nc1, not nc0, nc1. So what is nc1, 2 into nc2, 3 into nc3, etc. Till n into ncn, till n into ncn. Okay, let's find out this. They're only given the answer. So we have to just match our answer with that particular expression. But please understand here. Do not try to solve these kind of questions always by substitution of the value of n, even though that is a shorter way to do it. But unless and till, I mean, you are doing it in an exam where they'll allow you to do it. Always go always, you know, for a practice purpose, do it in a proper way. Okay. So how do you find this? How do you solve this question? Now, there are two ways to solve this question. One is by using your binomial coefficient properties which you have learned. And the second is, is by the use of calculus. So some of you are already use, some of you are already aware that we can use calculus for this purpose. So I'll start with the non-calculus approach first. Okay. Non-calculus means just by the use of your binomial coefficient properties. Okay. So how will it work? See, you're actually trying to find out, let's say I call this as s. Okay, let's say s n, whatever you want to call it. So your s n is nothing but you are trying to sum up r into n c r from r equal to, you can say zero to n, isn't it? It will not be wrong to start with zero, right? Because anyway, zeros, first term would leave when you put zero to be zero only. Okay. So how do you solve this question? Now, we have already seen that n c r is n by r into n minus 1 c r minus 1. Correct. So from here, can I say r into n c r is n into n minus 1 c r minus 1. Can I say this? So this term that you are having here, it can be further replaced with n into n minus 1 c r minus 1, r equal to zero to 1. Right? Yes or no? Now, please note, r equal to zero doesn't make sense over here because there's nothing like n minus 1 c minus 1. So you can actually write a one over here, which anyways did not disturb your summation value because anyways, r zero was a zero. So let's start with r one only. Okay. So it is going to be summed up from r equal to 1 till r equal to n. Correct. Now, write this n outside because your summation is applied to r, not to n, n is a fixed value, isn't it? So n will come out, it will not participate in your summation process, right? And you will be left with something like this. Okay. Now, think as if you are trying to sum n minus 1 c zero, n minus 1 c 1, all the way till n minus 1 c, n minus 1. What is this actually? What is this actually? This term 2 to the power n minus 1, isn't it? So please recall, we had done in our properties that if you sum up all the binomial coefficients from n c zero till n c, and it becomes 2 to the power n, instead of now n, if you put n minus 1, correct? This will become also n minus 1, correct? And that's where is your answer. This is your RHS. Okay. Hence proved. So this is your sum. Okay. But let me tell you, this process is not the most convenient way to solve the question. In fact, if given an option, I will not solve this question by this approach. But this is just to make you realize that this process is slightly cumbersome and there is a easier process waiting for you, which is our next process or next method to solve it, which is by the use of calculus. Okay. So let us go to that method. But if you want to note this down, please do so. I just write we know here so that we can, as when we refer to the notes, we can have a look at it. Okay. I personally do not like this method. Okay. So I'll be giving you a method number two, which is called your calculus method, non-calculus method and calculus method. Okay. So everybody has copied this down. Nobody has any question, any concerns. Okay. Great. So now again, in the calculus method, let's start with our given expression. This was given to us. By the way, again, I will also not write that n on the top, I'll directly write C0, C1 x, C2 x square, C3 x cube, datatata, datatata, cn x to the power okay. Now, here, if you see the question very carefully, you realize that these binomial coefficients are preceded by some number, for example, 1C1, if nothing is there, you can say 1C1, 2C2, 3C3, etc. Now, that number which is preceding this binomial coefficient are actually the powers of x that you see in this particular expansion. Now, which process in calculus helps you to bring down that power down. You will say it's a derivative. If you differentiate such terms, these powers are going to jump down and whatever will be left will be 1 less than that power, isn't it? So, here in light of that, again, this is something which has come from our observation. So, if you differentiate with respect to x on both the sides, the left-hand side will become n1 plus x to the power n minus 1, okay? This is going to become a 0. This is going to become 1C1. This is going to become 2C2x, 3C3x square and this goes on and on till nCnx to the power n minus 1, okay? Is it fine? Now, in order to make this x vanish from both the sides, put x as a 1. So, when you do that, on the left-hand side, you get n1 plus 1 to the power n minus 1, whereas the right-hand side will give you 1 times C1, 2 times C2, 3 times 3T. In short, it gives you the required series. That means the required series becomes n2 to the power n minus 1 and that's what we wanted to prove. That's what we wanted to prove. Is it fine? I personally find this approach or I personally feel that this approach is going to be much, much faster. So, when the problems become complicated, the first approach, the non-calculus approach will increasingly become more complicated. But this approach, the calculus approach basically is an easier way to get your answer. Is it fine? Any questions? So, which of the two methods you prefer using? The first one or the second one? I'm sure the second one is the one which most of you would like to use. Okay. Let's take more questions. Let's take more questions. Then we will understand the utility of the second approach. This will be the first one. So, in the last question we had taken C1 plus 2C2, this is a slightly different one. Here, you have to prove that. In fact, don't look at the result. Don't look at the result. Just look at finding the value of C0, 2C1, 3C2 till n plus 1Cm. So, everybody do this and let me know with a done on the chat box. Yes. Anybody? Analyze the expression first carefully. Then you'll come to know which approach is going to help you out. Okay. Harshita, very good. Anybody else? Okay. Arvind also done. Ashinta also done. Okay. That is one way to do it. See, again, I will use calculus approach here. Now, as you can see here, the coefficient of the coefficient that means the term which is along with the binomial coefficient is actually one more than the power of x that you have here. So, what I'm going to do? I'm going to first multiply this expression with an x. Okay. So, I'm just literally multiplying every term with an x. So, this will become this. Okay. Now, differentiate both sides with respect to x. Okay. Now, Harshita, what you're saying is when you remember the previous result, let's say this question comes as a standalone question. Will you carry that baggage of remembering those results before you solve this? Right. So, I'm solving it as if there was no prior art to this. Okay. Some result prior to this was not known to us. Okay. So, when you differentiate with respect to x here, please note that you're going to use the product rule over here. So, let's differentiate x first, keeping this way as it is. Then differentiate the second term, which is n1 plus x to the power n minus 1. And of course, derivative of 1 plus x will be 0, 1 itself. Okay. On the right side, you'll get c0, 2c1x, 3c2x square, till n plus 1cnx to the power n. Right. Now, put your x value as a 1 just to make the x vanish from the right side. So, that will give you 2 to the power n over here and this will give you n2 to the power n minus 1. And on the right side, you'll end up getting the required series. Isn't it? And on simplification, if you take 2 to the power n minus 1 common, it'll give you n plus 2. Okay. So, that is your right-hand side. Hence proof. Is it fine? Any questions, any concerns with respect to this? Okay. As it just is like as an exercise. Okay. For homework, try to solve this without using calculus. That means using a binomial coefficient series. Okay. Just try it for a homework. It would be a good exercise for you. Okay. So, try to solve this question without using calculus. For your January 30th monthly test, all these things will be tested there. Okay. All right. Should we move on to the next one? Okay. Let's try this one out. Again, it's a proof that question, but try not to look at the right-hand side. Think as if the question is asking you to find the sum of the series, C0 plus, C0 plus. Why does it come down? Something is not a problem. Yeah. C0 plus 3C1 plus 5C2, till 2n plus 1Cn. Yes. Any idea how to do this? Look at this term. This has a lot of things to say. 2n plus 1. 2n plus 1. Okay. Arvind. Good. Arvind thinks he knows. Okay. Try it out, Harshita. See whether it works. Whatever you are thinking or by the way, whatever anyone of you is thinking, please implement it. See whether it leads to your desired result. Okay. Done. Good. See here, I just like to bring your attention. If you want to double up this, okay, it's actually double of the power plus 1. Okay. So when you have been given this, you can do a small change here. Instead of X, write an X square. Okay. This will automatically double up the powers, isn't it? See, to the power 4, till Cn to the power 2n. Okay. Now, doubling up is not sufficient. It is important but not sufficient. I have to have a double up plus 1. So what I will do is multiply with X so that not only doubles up but adds a 1 to that doubled up term like this. Now bring that power down by differentiating with respect to X. Okay. So differentiate with respect to X both sides and because of that, so what will happen over here? See, just use a product tool over here. So 1 plus X square to the power n derivative of X is 1. Then keep X as it is. This derivative will be n1 plus X square to the power n minus 1 into 2x. Don't forget this 2x, right? Because derivative of 1 plus X square is 0 plus 2x. And on the right side, you'll end up getting something like this. Till 2n plus 1 Cn X to the power 2n. Is it fine? Any questions? Now you want the X to vanish. So put X as 1. Okay. So when you do that, the left hand side becomes 2 to the power n. 1 into n into 2 to the power n minus 1 into 2. C0 plus 3C1, 5C2, till 2n plus 1 Cn. Yes or no? So this is as good as n into 2 to the power n. Take 2 to the power n common and this is the result which we wanted to get. This is our right hand side. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Do let me know. Let's take this question now for a change. 1 plus X to the power of n is C0 C1 X, C2 X square, till Cn X to the power n. Then prove that. Then prove that this series is going to be this. Now there's a small change. Earlier the coefficients used to have those numbers which you see in the denominators multiplied to it. Now those numbers are getting and those numbers are basically dividing those coefficients. So think which process in calculus brings that power down. In fact, one more than that power down. And I'm sure you are aware of that process. Integration. Absolutely right Vishal. Okay. Anybody done? Okay. Now let me tell you something which the students normally do in the beginning. Okay. So let's say there's a student who does this. Okay. So he already knows that this is your given result. Now in order to get these numbers down, he integrates both side with respect to X. Okay. So I'm just writing that expression now by the way. So I'm assuming that you all have been introduced to integration to a certain extent in your physics. I know it's subject. Okay. So this is going to be something like this. Till CN X to the power n plus one by n plus one. Okay. Now, if I put a one on both the sides, put X as one on both the sides, you'll end up getting two to the power n plus one by n plus one. And on this side, you'll get C0 C1 by two C2. This is C2. Sorry. Not C3 C2. C2 by three, C3 by four, till CN by n plus one. Okay. Now I see here that in the answer, they don't have what I have got. They have got two to the power n plus one minus one by n plus one. Whereas I have got two to the power n plus one by n plus one. Where is that minus one gone? So what mistake have I committed over here? Can you repeat how to integrate? Okay. All right. So if you're integrating X to the power n, this answer everybody knows, what is the answer to this? X to the power n plus one by n plus one. Okay. I found this on the web for X to the power. Yeah. So if you have one plus X to the power n, what do you do? Very simple. You put one plus X as a T. That's another variable. So when one plus X is T, you differentiate both sides with respect to X. So your DX becomes DT. So this integration becomes one plus X to the power n, which is T to the power n. And instead of DX, you can write DT. So the approach is same. You'll end up getting T to the power n plus one by n plus one. And T is something which is one plus X. Okay. So this is how you end up getting. So that's why I wrote this result. One plus X to the power n plus one by n plus one. Fine. Any doubt with respect to that result? No. This result, which I have used. Okay. And I have put X as one. I ended up getting this. So what are the mistakes here, which I'm making? It's not a mistake in the, you know, this answer, but there is something which I'm missing out. I won't like to know from you. The constant of integration, exactly Satya. So the mistake which I made was I did not put a constant of integration, which ideally I should have put right after my integration. This is where I keep telling my students and to you as well that whenever you're performing an indefinite integration, do not forget this constant of integration. If you're taking this lightly, okay, because in 12th, when I do this chapter, many students are like, you know, sir, why do you put that C, you know, just waste our time and all. See, it's not a wastage of time. Your entire result will become faulty if you're not putting that C. Okay. So because of missing that C, my result did not come as expected. Okay. So things were wrong because I did not put my C value. So this whole thing is not going to make sense. Okay. Now, once I have put the C, I have to find the C value out. So for that, you put X value as a zero on both the sides. So when you put X value as zero on both the sides, you'll end up getting something like this, which clearly makes your constant of integration as negative one upon n plus one. Okay. So put this back over here. So your result becomes one plus X to the power n plus one by n plus one minus one by n plus one. And that is nothing but C zero X C one X square by two C two X cube by three data data data till CN X to the power n plus one by n plus one. Okay. Now here put your X value as a one. So when you do that, you big, if this becomes two to the power n plus one by n plus one minus one by n plus one, so you can take a minus one like this. Is it fine? And on the right side, you'll get C zero C one by two C two by three data data data till till CN by n plus one. Okay. And that is what we wanted to prove. An alternative approach here would be to integrate it from zero to one. Okay. So that is also another way to do it. I hope you know how to perform definite integral when the limits of integration is given to you. That could also be used. Okay. So if you don't want to put a zero, then a one, etc, you can do it in one shot. So right here, you can do it integration. So alternately, you can perform here. Let me write alternately. You can do the integration from zero to one of this expression. That also will give you a direct result one shot. You don't have to put a constant of integration there. There is actually called reverse chain rule. There's no chain rule. There's a reverse chain rule. And I think you were there in the class, which I've conducted during the bridge course. So those who are claiming that they don't know integration. Satya Manol, you were very much there in our bridge course class. Okay. So have a revisit of the bridge course class. Okay. You will, you'll get all the answers to the question that you're asking. So one plus x to the power n and all we all did there. Yeah. Is it fine? Any questions, any concerns with this approach kindly let me know. All right. Let's try one more where integration is involved. Okay. This is easy. Can you try this one out? Slightly complicated looking. So we have to find the sum of these series. In fact, we have to prove that or show that the sum of the series to square by one into two C zero. By the way, this is one into two, not one, one point two. This is one into two plus two cube by two into three C one plus two to the power four, three into four C two, two to the power n plus two C n by n plus one n plus two is equal to this expression. Everybody please try this out. Stopping in such a loud tone. Okay. See again, first forget about this two square, two cube, two four, et cetera that we can manage. How are we getting this one, two, two, three, three, four, et cetera? Right, Vishal, we need to integrate this thing twice. Okay. Okay. So we have already known this result. So I'll not be starting from scratch actually. I'll be starting from this result which we have already figured out in the previous slide. Okay. I think this was x, this was x square then C two by three x cube till C n by n plus one x to the power n plus one. Okay. Now in order to bring three here, four here, in fact, in order to bring two here, we need to integrate once more that everybody agrees. So this was the result. This was our previous result. Okay. And you need to remember this because you can't do the same obvious step again and again. Okay. So this was the result which we had done in our previous slide. So here now again, integrate with respect to x once again. Sorry, this was n plus one. Yeah. Yeah. Integrate this with respect to x once again till C n by n plus one x to the power n plus one dx. Okay. Okay. So let's see what happens to the left hand side. The left hand side will become, if I'm not mistaken, one by n plus one will be taken as a common integration of this is going to be, let me remove the integration sign because now we are just integrating. So it's one plus x to the power n plus two by n plus two minus x. Am I right? So see the constant term, I took it outside. Okay. It's like, you know, I brought this term. I brought the one by n plus one outside. Okay. Now I integrated this guy one plus x to the power n plus one, which is one plus x to the power n plus two divided by n plus two by the same logic. And minus one integration is x. Okay. Immediately put a constant of integration the moment you are integrating it. And this is going to become C zero x square by two. So can I write that two as one into two? Same here C one x cube by two into three. Then C two x four by three into four to C n x to the power n plus two by n plus one n plus two. Am I right? Now, before I put my C, I'm almost there. It's just that my x is to be now replaced with a two. Okay. But before I make that replacement, I need to first figure out what is this constant of integration for that put x as zero. So when you put x as zero on the left hand side, you'll end up getting one by n plus two. Okay. This will be anyways be a zero and this will be zero. Okay. Which clearly means your C is negative one upon n plus one n plus two. So put this back over here. Put that back over there. Okay. So this is going to simplify to one by n plus one one plus x to the power n plus two by n plus two minus x minus one by n plus one and plus two. Okay. So this is your C zero x square by one into two C one x cube by two into three till your C n x to the power n plus two by n plus one and plus two. Okay. Now put your x value as a two. Put your x value as a two. Okay. So when you do that, the left hand side will become one by n plus one times three to the power n plus two by n plus two minus two minus one by n plus one and plus two. Okay. That would be a required series. Okay. Now all I need to do is simplify this. So let us do that. Why to waste? So here what I'm going to do, I'm going to just take the LCM. So that is going to be three to the power n plus two minus two n minus four upon n plus one n plus two. And there is a minus one with the same base. So I can just do a minus one on top, which is going to make this as a minus five. Is this what we wanted to prove? Check. Three to the power n plus two. Yeah, three to the power n plus two minus two and minus five by n plus one and plus two. Is it fine? Any questions? Any concerns? Any questions? Any concerns? Okay. Good enough. Okay. So in the series, binomial coefficient series, we are now going to see next type of problem where you need some typical substitutions to solve them. So let me expose you to a few such problems. But before that, have you all copied this down? Any questions? Any concerns? Any cluster? No cluster? Okay. All unhappy? All right. So let's take another type of problems which require typical substitutions. Okay. Let's take this question. If one plus six to the power n is this, which is our regular series, no need to even look at it. Find the values of these series. Now, this time, I have not given you any answer to this. So it's a finding value. So find what is C0 minus C2 plus C4 minus six. So let's do the first one. Then we'll see the subsequent ones. Mind you, this is not C0 plus C2 plus C4. If it was plus in between, then I would have got just two to the power n minus one, which I already know. Okay. So here is an alternating sign also. And they're jumping from one. They're just leaving one coefficient and jumping to the other like that. So how do you do this kind of a problem? Do you want to try this out first? Please try this out. Take an attempt. See what to do. Think, think, think. See, we cannot give you all the types possible. So there will always be some type which will make you think what to do, how to do, let us think about it. Okay. See here, all of you. One very useful substitution in such cases is putting your x value as i. Yes. The same i that you came across in complex numbers. Okay. So complex numbers is not going to leave you. Okay. So let's put x as i. Right? We all know i, iota. Okay. So when you make that substitution, you'll end up getting something like this C0, C1i, C2i square is nothing but minus C2, then minus C3i, then C4, then C5i, correct? Then minus C6, dot, dot, dot, dot, dot, correct? Now, write the terms with i in them separately and write the terms without i, sorry, with i and without i separately. So this is what I'll be getting. Correct me if I'm wrong. Okay. Yes or no? Now, if you see question number one, let me call this as S1. Let's call this series as S1. Do you see that S1 is coming over here? And let's call the second question that is your, you know, second part of the question is S2. Do you see S2 coming here? So in one shot, I can find the answers to both first and the second part. How? Just comparing the real part on left and right, because this is acting like a real part of the left-hand side. And this is nothing but the imaginary part of your left-hand side. Correct? So if you find out the real part of 1 plus i to the power n and imaginary part of 1 plus i to the power n, you automatically get your series, isn't it? So how do you get the real part? Very simple. Now, everybody recall, everybody recall that you had already done that 1 plus i in polar form. How do you write it in the polar form? Route 2 cos pi by 4 plus i sin pi by 4. Everybody remembers that complex numbers? Please revise all these things because they are going to be tested. So if I raise it to a integral power by De Moivre's theorem, which theorem? Which theorem? De Moivre's theorem. I'm trying to, you know, pronounce it correctly. De Moivre's theorem, I can write it like this. Correct? In short, what I'm trying to say is that your left-hand side is 2 to the power n by 2 cos n pi by 4 plus i 2 to the power n by 2 sin n pi by 4. Okay? And that is nothing but s1 plus i s2. So your s1, that is nothing but c0 minus c2 plus c4, that is nothing but 2 to the power n by 2 cos n pi by 4. Right? Quite surprising that in the binomial coefficient, some you are getting a trigonometric ratio. Right? Yes, it is quite surprising. And if you compare your imaginary components, that is c1 minus c3 plus c5, it is going to be 2 to the power n by 2 sin n pi by 4. Now many people ask me, sir, is this result required to be remembered? In my experience, I have seen these type of questions being a part of a bigger problem. Okay? So if you happen to remember it, well and good. But if you don't happen to remember it, it's fine because such questions have come in the past, I agree. But they're not very, you know, frequent. That means the same type of question coming, let's say every third year or fourth year in competitive exams, that is very rare. But yes, they might come. Okay? So it is up to you. It is up to what your memory allows you. I'm not imposing any memorization here. Is it fine? Any questions, any concerns? Do let me know. Now having shown you the path to do such kind of questions, I would request everybody to try the third one out, which I think I have not done. Okay? So the first and the second part are done in this. The third part, I'm going to ask you to do it. So let me go to the next slide and post the question again so that we have enough space to work on it. But is everything clear over here? How to write this in polar form? How to write it? How to use De Moivre's theorem? Okay, because n is an integer over here. I mean, in fact, it's a whole number here. Whole number are integers only. So how to use De Moivre's theorem and how to just find out the sum of your series by using the comparison of the real part and the imaginary part. These things will not come to you from day one. You have to keep practicing, you have to keep applying, you have to keep thinking laterally. Okay? Okay, so if you have copied this down, time for us to do the next third part, which I'm again going to write it down. Where was the question? Where was the question? Where was the question? Let's do the third part. And I would give you some time to think on it and let me know if you have got a breakthrough. Anybody who has got a breakthrough in the third part, just let me know what you have done to solve that problem. Very good Vishal. I think you're on the right track. Oh my God, that's too complicated a substitution. See here, again, let's start with our given series. Okay, so this is C0, C1x, C2x2, C3x2. Let me write it till a few more terms. C5x would be power 5, C6x would be power 6. Okay. Now, putting one, we already know the result. Putting one, we already know the result. That is C0, C1, C2, C3, C4, C5, C6 till CN is going to be 2 to the power. Okay. So this is when you put X as 1. Now put X as omega. Okay, the complex cube root of unity. So when you put omega, you end up getting C0, C1 omega, C2 omega square, C3 omega Q, which is back to C3. Okay, C4 omega, C5 omega square. Okay, C6 omega 6 is back to C6. Then put X as omega square. Then put X as omega square. So when you make this substitution, you end up getting C0, C1 omega square, C2 omega 4. C2 omega 4 is C2 omega again, because omega to the power 4 is omega. Okay. Then C3 omega to the power 6, which is back to C3. This is C4 omega to the power of 8. So omega to the power of 8 is omega square. Okay. C5 omega to the power of 10, which is omega only. C6 omega to the power 12, which is C6 only. Okay. Now add the three series that you have written. Add the three series you have written, that means 2 to the power n, 1 plus omega to the power n and 1 plus omega square to the power n. See what you will see here that this will be C0, C0, C0 three times. Okay. 3 C0. Correct. This will be take C1 common. And you would realize that you will end up getting something like 1 plus omega plus omega square into C1. And this guy is a 0. So it will kill all the terms. Right. So this will be a 0. Same will be happening for this term also 0. Right. Because 1 plus omega plus omega square will come as a common term. And you have C3 C3 C3, which is 3 C3 again. Similarly, this will be 0. This will be 0. And you will have 3 times C6. In short, you end up getting 3 times the given series. That means your S is 1 third of 2 to the power n, 1 plus omega to the power n, 1 plus omega square to the power n. It's not a good practice to leave your answer in terms of omega and omega square, given that the entire series is basically consistent. I'm so sorry, given that this entire series is consistent of all, you know, non real terms. Right. So let us try to see whether this comes out to be actually a term which is real in nature. I don't want any omega, omega square to appear in our answer. So let us do us a small simplification here. See, we all know your omega value. Omega value is minus half plus i root 3 by 2. So 1 plus omega will be what? 1 plus omega will be nothing but half plus i root 3 by 2. As a complex number in polar form, can I write it as pi by 3 plus i sine pi by 3? So 1 plus omega to the power n, can I write it as cos n pi by 3 plus i sine n pi by 3? Anybody has any objection with this? Do let me know. No objection? If there is an objection, then objection overruled. Objection overruled, my lord. Okay, next. We already know omega square value. Let me write that first, then we'll add a 1. Omega square value is minus half minus i root 3 by 2. So 1 plus omega square will be what? Half minus i root 3 by 2. If I am not mistaken, this is actually cos pi by 3 minus i sine pi by 3. Am I right? So if we raise it to the power of n, can I say by deMauverie's theorem, this is going to become this? Agreed? Any objection to this? No objection? Okay. So if there's no objection, let us try to use that result over here. So it's 1 third 2 to the power n and this term is cos n pi by 3 plus i sine n pi by 3 and the other term is cos n pi by 3. Let me write it in white minus i sine n pi by 3. Now automatically you'll see the i related terms will get cancelled off thereby giving your answer to be 1 third of 2 to the power n plus twice of cos n pi by 3. This is your answer. Is it fine? Any questions? Any concerns? Do let me know. Clear? Happy? Now, if you are a J advance aspirant and let's say the question center gives you this question, he may put both the options in your option list. So be aware of both the versions of the result. Don't be like, okay, I will only know this. In fact, there's nothing to remember. It's all about figuring it out. Now, the next type of question which comes in this is called binomial within binomial. So we'll take that up and maybe then we'll take a small break because it's already six o'clock. So let me move on to the next type of question which is called binomial within binomial. Okay, now to understand this concept, I will be taking a small question. Okay, let's take a small question to begin with. Let's say I want to find out, I want to find out c0, I mean, when I say c0, nc0, it means, okay, minus nc1, 2 to the power n minus 2 cn plus nc2, 2 to the power n minus 4 cn again, okay, minus and this process goes on and on. Okay, let's say the question is find the sum of the series. So as you can see, the first term is nc0, 2ncn, okay, so these two terms are multiplied. Then you have minus nc1, 2n minus 2 in 2n minus 2 cn, then plus nc2, 2n minus 4 c4 and this continues on. Okay, how do you find this expression? What is this equal to is what we are looking for. Any idea? How do you solve this? Think, how can you get this result? Again, this is not meant for J main. This is actually a slightly heavier concept. Maybe they can ask you in KVP way, okay, or J advanced. But still, I would like everybody to think over it. How to solve this? See, first of all, few observations I can see. These are your normal nc0, nc1, nc2, etc. Okay, that is one observation that you have. Second thing that you see these terms have alternating positive negative signs. So there's a negative, then there's a positive, then there's a negative. Okay, this is another observation. Third observation I can see there is constantly nn over here. Even though the superscript is changing, in the subscript there is an nn. Okay, now in light of this, I have a small, you can say suggestion over here. Can I say this term, this term is the coefficient of, just the first term. Okay, coefficient of x to the power n, x to the power n, in 1 plus x to the power of 2n minus 1. I mean, let me just write it down for the first term. I will not write it for the other terms. So I'm just trying to say that this first term is the coefficient of, let me just remove this minus, is the coefficient of x to the power n in this term, nc0 into 1 plus x to the power 2n. Do you all agree with me on that or not? Okay, everybody agrees. The second term, let me write that also. The second term is the coefficient of x to the power n in nc1, 1 plus x to the power of 2n minus 2. Similarly, third term is the coefficient of x to the power n in nc2, 1 plus x to the power 2n minus 4 and so on. In other words, you are actually trying to find out coefficient of x to the power n in nc0, 1 plus x to the power 2n minus nc1, 1 plus x to the power 2n minus 2 plus nc2, 1 plus x to the power 2n minus 4, da, da, da, da, da. Correct. Agreed. Because this particular sum or this particular series will appear when you are looking for the coefficient of x to the power n in this. Now, pay attention here, very important observation I will show you in. Oh, yeah. So this term is nothing but all of you please pay attention. 1 plus x to the whole square to the power of n. Second term is nc1, 1 plus x whole square to the power of n minus 1. Third term is nc2, 1 plus x to the whole square to the power of n minus 2 and so on. Okay. Which means we are now finding. So see, everybody please pay attention. Take this term 1 plus x square to be like some number y in your mind. So it's like y to the power n, y to the power n minus 1. Okay. So can I say it is actually y, in this case y is our 1 plus x the whole square minus 1 to the power of n or not. Remove this nc0 not required anymore. Isn't it? So isn't this a binomial expression in itself correct? No. So imagine you treating this as some y minus 1 to the power n. If you write it, how will you write it? nc0 y to the power n minus nc1 y to the power n minus 1 plus nc2 y to the power n minus 2. This is how you're going to write. So imagine that this term is this way is actually written over here. So nc0 y to the power n minus nc2 1 y to the power n minus 1 like that. Okay. So in light of this, can I now reduce my problem to saying I have to find out what is the coefficient of x to the power n in this expression which is actually 2x plus x square to the power of n. Correct? Now if you take x out, you are actually finding the coefficient of x to the power n in this. Yes or no? Now x to the power n is already sitting over here. So you just require a constant from here that will give you the coefficient. So what is the constant term that you will get from here 2 to the power n? So this becomes your answer to the question because x to the power n is already there. So any x to the power n into that constant will be the term which will be giving you that x to the power n related term. So the coefficient of x to the power n will actually be the constant that you will get from 2 plus x to the power n which is only one constant 2 to the power n. See when you expand this, you get 2 to the power n, you'll end up getting nc1 2 to the power n minus 1x. So if you do a multiplication, x to the power n will occur only with 2 to the power n. Rest all the terms will have higher powers than n. So we'll have n plus 1. So the coefficient of x to the power n is this guy. That's what I wrote over here. So these kinds of questions are slightly difficult to crack. But don't worry, you'll be getting a DPP where you'll be having more questions of this type which you can practice. Okay. So we'll take a small break over here. I think this was a very heavy problem for you. So you need a break. So on the other side of the break, we will look into some more series and then we'll start with binomial expansion for any real power. That means it could be negative integer power or fraction power. And then we'll wind up this topic with multinomial theorem. Not the one which you did in PNC. In PNC, it was actually a resemblance to a multinomial theorem. But actually multinomial theorem was not used there. It looked, the expression looked like a multinomial theorem. Okay. Anyways, so let's have a break right now. The time as of now is 6.16. So we will meet exactly 6.31 pm sharp. Okay. On the other side, we'll take up a few more series on binomial coefficients and then we'll start with binomial theorem for any real index and then multinomial theorem. Okay. Enjoy your break. So this time, I'm going to talk about product of binomial coefficient. Let me write it like this. Sum of product of binomial coefficients. So this is another type of questions which is normally asked. Okay. Let me begin with a simple question here. There's no snapshot of that. Okay. Let's take this question. Find the sum of C0 square, C1 square, C2 square, C3 square. That means if you're taking the product of the binomial coefficient with itself, that means you're squaring those binomial coefficients and adding them, what is this sum going to be equal to and how are you going to find that sum out most importantly? Okay. Now, let me give you an idea of how to do these kinds of problems from this example and then I'll give you a follow-up question. Now, see, we already know 1 plus x to the power n. Right? What is it? C0, C1x, C2x square, C3x cube and so on till C and x to the power n. Okay. Yes or no? Right? Now, you write the same binomial term in a reverse fashion. That means put x as your first term and 1 as your second term. So, this is going to lead to this expression. Yes or no? Okay. Now, try to identify, try to identify that if you want to make C0 square, which two terms you need to multiply? Now, you will say, sir, obviously, sir, these two terms need to be multiplied. Okay. If you want to get C1 square, which two terms need to be multiplied, these two terms are needed to be multiplied. Okay. Similarly, C2 square, these two terms, C3 square, these two terms, dada, dada, dada. Cn square, these two terms are going to be multiplied. Right? Thankfully, C0 square, C1 square, C2 square, C3 square, etc., they all are actually coefficients of x to the power of n. As you can see, when you completely multiply these two terms, you will end up getting like C0 square xn. Isn't it? When you completely multiply these two terms, you will end up getting C1 square xn. When you completely multiply these two terms, you get C2 square xn and dada, dada, dada, till this term. Now, what is this actually? The observation here is that your given series, let's say I call this as s, the given series actually occurs as a coefficient of x to the power n when you multiply these two binomial expressions. Right? So, what I want to say here is when you multiply these two binomial expressions, you will end up getting several terms. In that term, they will be having something like x to the power n, dada, dada. By the way, please note that the maximum power it can go up till is x to the power 2n. Okay? So, in that particular series, when you are multiplying, you will get one such term x to the power n whose coefficient will be your given series. Isn't it not? Because, see, when you are multiplying the top series with the bottom series, you will get Kethorov terms right from constant till x to the power 2n. So, I am not interested in all those terms. I am interested in one particular term which I identified as x to the power n because that x to the power n coefficient is basically these expressions which I have in my series. Are you getting my point? So, in short, what I want? In short, I want the coefficient of x to the power n in this expression. So, in this expression, I just need what is the coefficient of x to the power n that itself will be my required series. Are you getting my point? And what is the coefficient of x to the power n in 1 plus x to the power 2n? Is 2n cn? So, this becomes your answer to that question. Is it clear? Right? Any questions? Any concerns? So, here the standard operating procedure is, we try to figure out which consistent power of x is coming along with all these terms. Okay? So, here x to the power n was occurring with c0 square, c1 square, c2 square, c3 square. So, in totality, the coefficient would have been the sum of all of them, which is actually your series. So, if I am able to find out what is the coefficient of x to the power n in the product of these two, that coefficient will be my required answer. Are you getting my point? Now, many people think that multiplication happens one below the other. No, I am not adding. I am multiplying. Multiplication happens one term with every other term. Okay? So, I am not interested in all the terms. You will end up getting almost several terms. I am just interested in one particular term, which is x to the power n. Are you getting my point here? Now, many people actually remember this result also because it is useful in solving some complicated version of the question. Is it fine? Any questions? Any concerns? Okay. Try this one out. Can I move on to the next slide? Find the sum of this series, c0, c1, c1, c2, c2, c3, till cn minus 1, cn. Try this out, everybody. Let me see whether you are able to extrapolate that concept to solve this question. Yes. Let me just write down our regular series here. Let me write the same series in a reverse order also. How is that possible? Now, all of you, please pay attention. Yeah, that is correct. See, this term will be obtained when these two guys will interact. This term will be obtained when these two guys interact or there is a multiplication happening between these two terms. Similarly, this term will come. c2, c3 will come and these two guys will multiply. Now, what do you see here is that? I mean, just let me write. Yeah. What do you see here is that this term or this term or this term or this term or till the last term actually is coming with the x to the power n minus 1 power. Do you see that? So, if you do a complete multiplication, c0, c1 comes along with x to the power n minus 1. Similarly, c1, c2 also comes with x to the power n minus 1. Yes or no? Likewise, the last term also that comes along with x to the power n minus 1. Yes or no? In short, if you multiply these two series, if you multiply these two series, you will end up getting so many terms. But I am interested in that term which comes along with x to the power n minus 1. And that term is what will be carrying your given series as your coefficient, isn't it? So, the problem boils down to finding the coefficient of x to the power n minus 1 in this expression, which is nothing but 2n cn minus 1 which you can also write as 2n cn plus 1. Both are fine because of your ncr equal to ncn minus r property. Now, some of you would have done this. Some of you would have multiplied these, isn't it? Let me put them in white. This, this. In that case, you will have to see what is the coefficient of x to the power of n plus 1. And you will be surprised to know even x to the power n plus 1 will have the same series as its coefficient. So, either you do like this or you find the coefficient of x to the power n plus 1 in the product or in 1 plus x to the power 2n, which ultimately gives you this expression. And as you already seen that they are both equal. So, either of the two mechanisms can lead to the same result. Okay. Now, let me generalize this. Let me generalize this. But before I generalize it, please copy anything you want to. And if you have any questions, please do let me know. Okay. So, I'll just generalize this. In case this result is quite useful to you, you can use it. So, the generalization is if you have c0, cr, c1, cr plus 1, c2, cr plus 2, tils cn minus rcn. Okay. This result is given to you as 2ncn minus r or 2ncn plus r, both are fine. Okay. So, please remember this. So, this might be helpful to you in solving a complicated version of the same question. Okay. So, in our first example, our r was 0. Isn't it? That's why we got c0 square, c1 square, c2 square. That time the result was 2ncn. In our question, which I gave you on this slide itself, your r was 1, isn't it? c0, c1, c1, c2, c2, c3, etc. So, your result became 2ncn minus 1 or 2ncn plus 1, both are fine. Okay. So, this is something which might be helpful to you in solving questions. Okay. So, with this, we are now going to go towards binomial expansion for any power. Okay. That's a simple result. I'll just be giving it to you. Binomial theorem for any index, for any index. That means any real number power. It could be fraction. It could be negative integers. Right? So, let me start with a very simple expression, 1 plus x to the power n, n being any integer, sorry, n being any real number. So, it could be negative integers also. So, in this, in these cases, you realize that your expansion goes like this. I think I've already discussed this in your bridge course also. So, here, couple of things to be noted down. Number one, it's an infinite series. It's an infinite series. If your n is not a whole number, why? Because in our whole number indexes, we used to stop somewhere because we used to reach a stage where we have to write n minus n, isn't it? See, actually speaking, when you write 1 plus x whole cube, let's say whole square. So, you have 2c0, 1 to the power 2 plus, okay? In fact, 2c0 is nothing but 1 itself. Then you have n into, n into 1 to the power of 1, x to the power 1. Then you have n into n minus 1 by 2, x to the power, sorry, 1 to the power 0, x to the power 2. And the next term, which ideally we don't write, is 2 into 2 minus 1, 2 minus 2 by 3 factorial n. Okay, whatever comes in. But this entire thing goes for a 0. Sorry about that. So, that is why this particular series comes to a stop, comes to a halt. But such a thing is not going to happen if your n is not a whole number. It's going to go on and on and on forever. Are you getting my point? Because you're always subtracting an integer from that number. And if that number is a negative integer, it is never going to vanish. If it is a fraction also, or let's say a number having some decimal values also, that is also never going to vanish. That's the whole reason why this entire series goes on and on forever. Okay, that's number 1 point to note down. Number 2 point to note down, we do not use, do not use ncr if your n is a non-whole number. We should never use ncr expression. There is nothing called minus 1c1. There's nothing called 3 by 2c1. Okay, that is why instead of nc0, nc1, nc2, we are actually using the expansions of it. Okay, third thing is for this series to be convergent or for this binomial theorem to hold true, your mod of x should be less than 1. This is called the condition for convergence. I'm nothing new. This is something which we have already had a taste in our bridge course also. Now, why is this expression mod x should be less than 1? That means why should my x be between minus 1 to 1? The reason for that is if you see the right hand side is an infinitely term expression and your left hand side is a finite term for a given x. Yes or no? For a given x and for a given n, your left hand side will be a finite term unless n del x itself is very, very large. And n is very, very large. So for a finite value of n and a finite value of x, your left hand term will be finite. But your right hand side term may become infinity because you're summing up till infinity. You're summing up till infinitely many terms. So to keep your left hand side and right hand side values coherent with each other or agreeing with each other, we need to ensure that even if the right hand side is going up till infinity or infinite number of terms, its sum should be finite. In other words, you want that particular series on the right side to be a convergent one. Convergent means even though if it is summed till infinite terms, its sum should be a finite value. And for that to happen, this condition is fulfilled. Now don't ask me why this condition came. It is going to be taught to you in your second semester of your undergrad when you are going to learn series chapter there. So there are various tests for convergence. Okay. There's a limit test. There is a ratio test. There's a Rabi's test. There's a DLM word test. There are so many tests. There's a P, there's a P test. There's an integral test. So all those tests you're going to study in your undergrad not right now. Okay. So these are the three things you need to keep in your mind while you are writing a binomial expression for an infinite term. Okay. Now many people will ask me, how will it change if there was an A over here instead of an A? Okay. One I took for my convenience but if there was an A there, what would you do? And what would be the radius of convergence? See, nothing different if you took, if you had an A over there, you could actually write this as this. Okay. And you can again use your same binomial expansion that we had discussed here. It's just now your X will be X by A. Nothing else will change. It's just that now your X will be X by A. Okay. So what will be your condition for convergence in this case? So condition for convergence here will be, this term X by A mod should be less than 1. That is to say mod X should be less than mod A. Okay. Let's say A will be given to you as some number. So please take that into your consideration. This will become your condition for convergence. Is it fine? Any questions? Any concerns? Let's take few questions based on the same. Okay. Let's take this question. Assuming X to be so small that X square and higher powers of X can be neglected, prove that this expression can be simplified to this. Okay. In fact, they should use the symbol. That means it is approximated as 1 minus 3.5 by 96 X. As you can see, each of these factors, they are having some kind of a non whole number powers on that. So see, there's a minus four. There's a half. There's a two-third. Done? Okay. Let's look at this term. Okay. Let's see if I want to approximate it. So this is acting like my X of the formula, right? So it's 1 plus n X. Okay. The rest of the terms don't bother to write because as I, as they already told you, X square and higher powers of X can be neglected. So this is approximately equal to 1 minus 3 X. Okay. Look at the other term. So here also, I'll do one small activity and you know, correct me if I'm mistaken here. I'll take the 16 out. Okay. When 16 comes out, it will come to the subject of the power half, which means I'll get something like this. Now, many people ask me, sir, why do you do like this? Like, because he having a one is always a convenience, right? Because one, you don't have to worry about what power is there on one. Okay. You have to only worry about the next term. Okay. So this can be written as four times one minus. Please note that it is actually this one plus half into minus 3 X by 16. Okay. Which is approximately four times. This is one minus 3 X by 32. That's nothing but four minus 3 X by approximately, approximately, as per the given conditions given to me. And likewise, the last term, which is a denominator term, which is this, if I take eight outside, eight to the power two by three will be four, if you all agree with me. Correct. So this can be approximated as four times one plus two X by 24. Can I write it as X by 12 to be more precise rather than writing 24 and all. So this is approximately four plus X by three. Okay. Now, let's put it in the given expression. So then the top two terms will be multiplied. So our given expression could be approximated as one minus 3 X four minus 3 X by eight. Okay. And see when you're taking it to the numerator, it just becomes four minus X by three. Why does it become four minus X by three? Because let's say there was a power of okay, let me just do it over here. First of all, then I'll write down the result not directly. Yeah. So let's say this you took it on the other upper side, it would be a power of minus two by three. Right. So minus two by three will give you minus two by three will give you one upon four. Sorry. So one upon four will come up. Sorry. Sorry. Yeah. So this will be one upon four and this will be a minus sign. Yeah. So this is going to give you a one by four minus X by 48. Okay. So that will get multiplied over it one by four minus X by 48. Okay. Now just focus on constant terms over here. So constant terms will obtain when this this and this multiply, that is clearly a one, then focus on X terms. X terms will come and this multiplies with this and then multiplies with one by four. That's one way to get a constant term. Correct. That's going to be minus three X. Correct. Or this minus three X by eight multiplies with one here and one by four over here. That will give you minus three X by 32. Or this term multiplies with four here and one over here. That's going to give you that's going to give you minus X by 12. Okay. Let's try to see how much it comes out to be. Okay. So let's take a LCM of of 96, I believe. So this is going to be negative 288 minus nine. Correct. Minus eight. So that's nothing but that's nothing but one minus three, not five by 96 X. Okay. Is it fine? So very important tool when it comes to approximating things like this. So very, very useful in physics chapters also. Is everybody clear how this works? So what I did first was I took this expression. Let me write it once again here. I took this expression like this one plus one plus three X by four to the power minus four, 16 minus three X to the power half. And I wrote it as eight plus X to the power minus two by three. Each one of them I approximated. So as you can see, I've approximated each one of them. And then I finally multiplied it. And here also I'm only interested in constant terms and coefficients of X. Higher powers of X, including X square are all neglected. Okay. So this is how you end up getting the given result. Is it fine? Any questions? Any concerns? You have, we want to have a look at this expansions. Please have a look at it. How I've done it. Any questions? Anybody? No questions? Chalo great. So with this, we move on to another question. Find the sum of the series to infinity one minus one by eight plus one by eight into three by 16 minus one by eight, three by 16, five by 24, and so on and so forth. Yes, sir. Anybody? Anybody with any success? Okay. See here everybody, please pay attention. You have to be a very good observer over here. You have something like one by eight. Okay. Now this time that you have one by eight and three by 16, let's say I take an eight common. Okay. I'll get a one into two, which is actually two factorial. Correct. Now see, I have this particular expression in my mind. The binomial expression that we have written. Okay. So one plus nx, n into n minus one by two, sorry, n into n minus one by two factorial X square and so on. Correct. The next term will be n into n minus one, n minus two by three factorial X cube. Okay. Now let's see. So this two factorial that is sitting over here, I am able to make it. Okay. Now one and three, the gap is two. So when you subtracted one from the first expression, you end up getting a three on the top. This makes me believe and correct me if I'm wrong. This makes me believe that there was a one by two here. Okay. In fact, minus one by two here. Again, and you subtracted a one from here, which gave you minus three by two. Okay. This was two factorial and I think one more two is left off. Okay. By the way, a small mistake which I'm doing here, it would be eight square because if you're taking eight, common from both of them, it will be eight square. Okay. So I have to have a 64 here. So this will also change once again. So this is 64. Correct. Now 64 and you've already taken out two, two from here. So four has been taken. So 16 and 16 should be square of something. So can I say it is something like this? So this term gives me a hint that it could be of this nature. Am I right? That means I got an idea about the first term also. The first term is actually minus half into one by four. Right. Now it is making sense to me that my next term should have been minus half minus three by two minus five by two by three factorial one by four cube. And you realize that this expression is your given term over here, this term. Check. Of course, with the minus sign. So let me in gross the minus sign also. Check it out and let me know whether you are convinced. Right. So you have to be a good observer here. You can't solve this question blindly. Okay. So this is actually, this is actually the expansion of one one plus four to the power of minus half. That means this entire series comes out to be this, which is actually four by five under root, which is nothing but two by root five. So the answer to this question is this series is going to give you answer as two by root five, plain and simple. Over. Question solved. But easier said than done. This finding this entire expression is not that simple a task. Many people falter there. Okay. Is it fine? Any questions? Any concerns? So when infinite series is given to you, please keep binomial theorem for non whole number powers also in your purview, also in your consideration. So all infinite series are not inspired from Maclaurin series or infinite series are not expired from VN method. So there could be some infinite series, which you could solve by using your binomial theorem for non whole number index. Is it fine? Again, to build up this, you have to practice questions. Okay. Don't worry. We'll be sharing the required DPPs and I'm sure some DPPs have these type of questions. Okay. So in the interest of time, we only have 25 odd minutes left. So we'll be quickly discussing multinomial theorem just to wind up this chapter. So our next topic will be multinomial theorem. So multinomial theorem is a theorem where you will be learning. In fact, we get to know how to expand these kind of expressions. Okay. Where instead of two variables or two, you can say dissimilar terms, we have our dissimilar terms. And again, you're raising it to a whole number power. Okay. So we are only going to restrict ourselves to whole number powers. We are not going to talk about how to use multinomial theorem when the power is a non whole number. So that is beyond our purview. We are not going to study about that. So this expression again, again, I'm going to give this to you without any proof. This expression gives you summation. Note this down expression. I'll just give an example to illustrate this expression. So it's summation n factorial upon alpha one factorial alpha two factorial till alpha r factorial x to the x one to the power alpha one x two to the power alpha two to the power alpha r. So it's a summation of this where where alpha i's are all whole numbers such that their sum is their sum is equal to n. I know there's a very ugly expression to look at. But I'll just give an example to illustrate this. First note this down. And I'll also tell you why did not I mention any upper lower limits of summation because it depends upon your n value. So we'll discuss about that. It depends upon your n and r. Okay. So first note this expression down, then I'll give you a simple example. This is r only. Okay. So sorry for that overwriting there. Okay, done. Okay. So to make you understand this formula, which I have written, which quite looks quite ugly, I'll give you a very simple example, which you probably will be aware of a plus b plus c the whole square. Okay. Now the reason why I've chosen this expression is because it's a multinomial. First of all, because ABC three terms are involved. And secondly, you know the expansion of it. Right. So that when I write it, when I finally write down my result using this formula, you can match it with the result, which is already known to you. Okay. So that you are convinced that this formula works. Now as for this formula, we have to first write down three such numbers. In fact, our hair is three. So we have to write three such numbers, alpha one, alpha two, alpha three, which are actually whole numbers where their sum is going to be N. N is two in this case. Okay. Now let us list out what all possible values of alpha one, alpha two, alpha three can come up such that their, their whole numbers and their sum should be equal to two. So it's very simple. One combination will be two, zero, zero. Correct. So they're all whole numbers and their sum is two also. Another would be zero, two, zero. Let's go step by step. Then zero, zero, two. Okay. Then one, one, zero. Then one, zero. In fact, let me write it like this zero, one, one. And then last combination will be one, zero, one. Is it fine? Is there any other possibility? No possibility. Okay. Now let us make use of this summation process. So it is summation of all these kinds of terms with various combinations of alpha one, alpha two, alpha three, which you have already figured it out. Okay. So the first sum that I will be getting is N factorial, which is nothing but two factorial alpha one, alpha two, alpha three's factorial. So as per my first case, as per this case, it will be two factorial, zero factorial, zero factorial. Then x one to the power alpha one, which in this case is a, a to the power alpha one x two to the power alpha two, which is b to the power zero and x three, which is our c, c to the power zero. And keep writing summation of all such terms. So the next term will be again two factorial. Now focus on this zero to zero. So zero to zero means zero factorial, two factorial, zero factorial, a to the power zero, b to the power two, c to the power zero. Again, for the next term, focus on this third one. So we have covered up this one. Now focus on this third one. So as per the third one, my expression will be two factorial, zero factorial, zero factorial, two factorial, a to the power zero, b to the power zero, c to the power ten, sorry c to the power two. Then focus on the fourth one. This is our fourth one. Alpha is one, sorry, alpha one is one, alpha two is also one, alpha three is zero. So it will become two factorial, one factorial, one factorial, zero factorial, a to the power one, b to the power one, c to the power one. Okay. Again, the next term will come from this guy. Okay, zero, one, one combination. So zero, one, one means it will be two factorial. I am changing the pen color. Yeah, two factorial, zero factorial, one factorial, one factorial, a to the power zero, b to the power one, c to the power one. and the last term will come from this combination okay so that is going to be two factorial one factorial zero factorial one factorial a to the power one b to the power zero c to the power one now this term itself when you simplify you are going to see a square as you can see this is going to be a square this is going to be b square this is going to be c square this is going to be 2ab check this is going to be 2bc and this is going to be 2ac or 2ca okay so the expansion that you use to write effortlessly in your class 10th or 9th whenever you learn that is actually based out of such expansions okay now don't worry nobody's going to ask you to expand multinomial theorem for sure but what you should actually figure out is this expression okay so they may they might ask you the coefficient of certain expression in a multinomial theorem so please remember this this is very very important is it fine any questions any concerns that you have with respect to this okay I think okay so just remember this expression no proof and all is required for this let's take questions for the next 15 minutes we'll take a few questions based on the same let me start with very simple question how many total number of distinct or dissimilar terms would be there in this expansion so this takes me to our first class the very first session that we had on binomial theorem where I talked about how many dissimilar terms are there in a multinomial expansion like this does anybody remember that so maximum number of dissimilar terms okay how much is it so if let's say you had x1 x2 xc let's say they all different from each other okay then the maximum number of dissimilar terms used to be n plus r minus 1 cr minus 1 right so in our case in our case the answer would be n is n r is actually 4 so it would be n plus 4 minus 1 c4 minus 1 which is n plus 3 c3 so this will be your maximum number of dissimilar terms okay now the question actually doesn't say maximum number but see here the problem is the the question is basically dealing with such terms which are dissimilar themselves so they're assuming that the product of two dissimilar terms is not creating you know a term which is obtained by the product of the other two that means no combination of powers on x, y, z and w is creating an expression similar to any other combination are you getting what I'm trying to say okay so basically they're trying to say see let's say if I say 1 to the power 5 or 1 to the power 2 okay 1 to the power 2 can also be obtained when we have something like this also multiplied correct so if you had 1 plus x plus 1 by x okay they raise to certain power let's say power of 2 so if you have an expression like this or if you have an expression like this they will both be of similar nature because both creating are the both are producing constants here okay so in that case your formula is going to fail that is why I wrote maximum number of dissimilar terms will be this which the question setter is assuming that all the terms will be dissimilar okay so that's an assumption that they have made but please understand please understand it actually gives you the maximum dissimilar terms there's something which they don't write on the screen here sometimes I scribble and it doesn't write yeah is it fine okay so we'll move on to the next question find the coefficient of a to the power 4 b cube c square d in this expansion or you calculated it also Satya very good I mean you can leave your answer in terms of factorial I don't mind it okay so first thing that you would observe here is that this sum should add up to 10 okay so there is some power of a some power of b sorry some power of minus b some power of c and some power of minus d which is actually resembling this expression okay and it will be preceded by let's say I call these numbers as alpha 1 alpha 2 alpha 3 alpha 4 so it will be having a coefficient which is 10 factorial by alpha 1 factorial alpha 2 factorial alpha 3 factorial alpha 4 factorial okay now if you try to make a direct comparison alpha 1 comes out to be loud and clear right as 4 now alpha 2 ideally should come out to be 3 but that would create a minus sign but let's wait because minus sign is in the last term also so as per me alpha 2 should be 3 alpha 3 should be 2 and yes alpha 4 should be 1 and thankfully because there is an odd here odd here minus minus what adjusted okay so this term is what we are going to okay this is what we are going to get as a coefficient of that given term okay so whatever it comes out to be I think some of you have figured it out but this is what I needed is it fine any questions any questions all right let's take this question find the coefficient of a q b 4 c to the power 5 in the expansion of b c plus c a plus a b to the power of 6 okay hashita okay one minute more to solve this I got answers for only two of you satyam and hashita so far what any uti and all is going on in school achcha chanta is not there because of that semester two uti two okay which uti tomorrow came on Thursday and people are absent today wow it's like you have to go to a party day after tomorrow you have stopped eating from today of course it has to be a holiday tomorrow right tomorrow is ganthan to divorce right republic did isn't it okay six factorial okay anybody else okay so let's say that particular term that you're looking for has got alpha 1 alpha 2 alpha 3 as the powers of b c c a and a b respectively now two things to be kept in mind number one alpha 1 alpha 2 alpha 3 should add to 6 okay and number 2 so if you see here your powers of a would be alpha 2 plus alpha 3 and that should match with 3 powers of b will be alpha 1 plus alpha 3 and that should match which 4 okay and powers of c will be alpha 2 plus in fact alpha 1 plus alpha 2 which should match with 5 okay now whichever alpha 1 alpha 2 alpha 3 satisfies these two set of conditions simultaneously those will be put over here and that will be my answer okay now but very very easy to find out so from here I can say alpha 1 is going to be 3 from here I can say alpha 2 is going to be 2 and from here I can say alpha 3 is going to be 1 okay so your coefficient of that required term which is a cube b 4 c 5 is going to be 6 factorial upon 3 factorial 2 factorial 1 factorial so it's 720 upon 6 into 2 which is 60 60 is going to be your answer is it fine everybody has got 60 those who said done and those actually some of you gave me the answer so well and good so here many people do a small mistake they think are you the sum is not coming out to be 6 by the way the sum this a b c are not your multinomial terms here multinomial terms are b c c and a b so don't directly try to compare the sum of the powers with this power 6 that is not going to match okay so this is a question with a slight twist is it fine any questions any concerns so we are not going to stop here because you know we have done enough of this chapter okay