 binomial theorem. Okay, this topic is not a new topic to you because I had already exposed you to this topic when I was doing limits concept with you. Remember, we had one standard algebraic limit x to the power n minus a to the power n by x minus a extending to a as n a to the power n minus 1. If you remember that limit, that was actually derived by the use of binomial theorem. Okay. And you'll find the use of binomial theorem in subsequent parts of your learning in physics, chemistry and maths. Okay. So let's talk about it. But before that, let me give you an overview of the chapter. Okay. What are there in these chapters in this chapter? So first, we're going to talk about binomial theorem applied to whole number index. Okay. Now, what's the binomial expression? Most of you are already aware of it. So binomial expression is where you have an algebraic expression containing two dissimilar terms raised to an index or an exponent. Okay. So what's the binomial term? Binomial term is nothing but it's an algebraic expression. Okay. So this is an algebraic expression, having two dissimilar terms, dissimilar terms raised to some exponent. This is called the exponent or index, exponent or it's also called as index. Okay. So we are going to learn in this chapter how to deal with such kind of an expression. Of course, binomial theorem is one aspect of this. There are many more things into the into the into the chapter. So binomial expression is this, but but let me tell you guys and girls, many people have raised this question to me that but binomial theorem can be applied even if these two terms are similar, right? Or even if these two terms are non-algebraic. Yes, absolutely right. This is a binomial expression. That doesn't mean binomial theorem is only going to work on a binomial expression. It can work on a non-binomial expression also. That means even if there's this expression has got two similar terms or even if these terms are non-algebraic in nature, let's say they're logarithmic or exponential, you can still use binomial theorem on that. Okay. So don't get confused between where binomial theorem can be used and what is called a binomial expression. Okay. Anyways, so our first part of the chapter, maybe we will not be able to complete the first part only in today's session. Okay. So this is going to be almost your 50 percent of the topic. So binomial theorem where your index is a whole number. This is what we are going to discuss in the first part of the topic. In the second part of the topic, we are going to talk about binomial coefficient series. Binomial coefficient series. Okay. This is what we are going to discuss. This is very important and especially important from your J main and J advance point of view because many questions have been framed. I've also seen many questions coming in CT also based on this concept. Binomial coefficient series. The third thing that we are going to talk about is binomial theorem for any index. Binomial theorem for such index, indices, which are non-whole numbers means it could be a fraction. It could be a negative integer. Okay. So we'll be talking about that as well and we'll also be talking lastly about the multinomial theorem. So this would be our four major subtopics that we are going to discuss under binomial theorem. Okay. Now enjoy this chapter. Most of these concepts are already explained to you directly or indirectly in our previous discussions and this is going to be our last topic. Two classes will be required. So I think the class that we will have on 26 January happens to be our next class. Are you ready for a class on 26 January? How many of you want class on 26 January? 26 January, will you attend a class of mine if I keep it? Of course, morning time is free because let's say if you are supposed to participate in some school event and all you can do it. Okay. We'll see when we... Oh yeah. Many people are now at home. Only school and all is has been closed off. Okay. Let's see. Let's see. 25th. We will decide what to do. Okay. Normally Tushar sir is going to poke me. Why are you keeping a class on a national holiday? This is the last class of this year. Why? This year. This year has already started by the way. You're talking of this academic year? No, no, no. The one more class is required and we'll have revision after that. So till your exam is over, we will not call it as the last class. Okay. One more thing I forgot about it. Please share with me your final semester schedule. When is it starting? When is it ending? Okay. So that we can, you know, stop at least five, six days before the exam. If you've got it, please somebody share it to me. Share it with us on the group so that all the teachers can have a look at it. Okay. So let us try to start. Let us start with binomial theorem for the first part. Binomial theorem for whole number index. Binomial theorem for or whole number index. Now I'm going to derive this theorem merely by observation. Okay. So whatever we have done in our junior classes, based on that, I'm going to derive the theorem which will help us to expand a binomial expression. Okay. So we are not talking about how to expand such kind of an expression where n is a whole number. Now let's start with our very, very simplistic case where n is a zero. Okay. So we all know that this is going to be a one, isn't it? So that one, I'm going to write it as one x to the power zero, a to the power zero. Okay. Just to make you realize some patterns. So I'm writing it in a very strange way. Most of you would be thinking, sir, one, you are writing it like this. Why? Right? Because I want you to realize some pattern which you will eventually realize when I, when I write the expansions of few more expressions like this. Let's say I want to write this, right? This is x plus a only, but that x plus a, let me write it like this. X plus a whole square. We all know it's x square plus 2x or 2x a plus a square, but I'll write it in a fancy way. Okay. See guys, pattern, pattern is what you should be paying attention to. What is the pattern that you are seeing over here? That's what is important. A x plus a whole cube is x cube, x cube, I'm writing it like this. 3x square a, 3x a square, and finally a cube, which I'm going to write it like this. Okay. So I can continue on and on, but what I want you to observe here is with respect to these expressions, what is the pattern that you are seeing? The ones which I'm underlined. What pattern do you observe? So one pattern that everybody would have seen is that each of these expressions have both x and a terms, right? So whatever terms I've written, it has got both x and a terms. Okay. And this exponent is first completely taken by x. So x raises itself to that exponent. A starts with zero and slowly, slowly x starts losing that exponent right? And a starts gaining. So here, if you see this, I mean, zero, there's nothing to see. But here, if you see in the second expression, one has been taken by this guy completely, is having zero and slowly x becomes, I mean, one less, it loses down one of the exponent and a gains one exponent. Okay. And at every time we realize that the sum of the powers of x and a matches with this number. Isn't it? Okay, let's see the second one also. So the exponent is two here. So two is completely taken by a first, a zero, then one for x, one for a, then zero for x, two for a, and at every stage, you'll see that the sum is going to be two. It's not going to be any other number. Okay. Same goes with here also, three also, three, zero, two, one, one, two, zero, three. Right? So it basically gives you a feeling as if you are trying to distribute the exponent between two people. Let's say there are three chocolates. Okay. And x and a are takers for those three chocolates. Let's say you want to distribute three chocolates. Okay. And they're two students x and a. So how many ways can you distribute it? Three to this guy, zero to this, two to this guy, one to this, one to this guy, two to this. And I mean three chocolates with three identical chocolates. Okay. Right. So basically these become the way you are distributing these chocolates to x and a, these basically become the way you are distributing the power n over x and a over here. Are you getting my point? So thanks to this particular observation that at least now if I have to write a further more expansion, let's say x plus a to the power four, at least I can say there'll be something x to the power four, a to the power zero, something x to the power three, a to the power one, something x to the power two, a to the power two, something x to the power one, a to the power three, and something x to the power zero, a to the power four. Correct. If I have to further write down x plus a to the power five, I can say something. I mean, what is that something and all? I'll fill that bracket in some time. Don't worry about it. So I'm just trying to come from the observations that we are, we are, you know, getting from the previous expansions. By the way, UTs are over. Can you share your maths marks with me? If at all, you don't mind? I think most of you were busy last week with your UTs, right? If I'm not mistaken, HSR, Kormangala, UTs were on. Very good. Pratej, it is out of how much? Excellent. That's good marks. How about others? And PSHSR hasn't got yet. What about Kormangala? Tell, tell. Don't hide. Shardili. Vashnav. Vashnav is not there. Karthik Sanovich. Message me privately. I will not reveal your marks. Sethu is also there. No, Sethu. Yeah, Sethu. Sethu is very quiet nowadays. Aiyo, Rama, Krishna. But still, how much you got? Tell, tell. Your class is waiting. Okay, good. That's good, Sethu. Shardili, good. I mean, you can always do better. Okay. See, 11-th is all of a game of speed, accuracy. Okay. Intelligence is definitely something which is required. But as you practice and you solve more and more questions, you work on your accuracy and speed. That is more important. Okay. Anyways, so coming to the pattern that we were talking about, the pattern is more or less obvious when it comes to these expressions which I have shown with a yellow underline. But what about these numbers? What about these numbers? How are they obtained? Okay. Now, long, long ago, there was a French mathematician, Blaise Pascal. Okay, Blaise Pascal. I mean, in our Indian curriculum, we call him Pascal. So Pascal. See, many times I also, you know, call him Pascal only. Aiyo, that's what happens. You have to check the paper, no, Sethu. Okay. So long ago, there was a mathematician by the name of Blaise Pascal or Pascal. He came up with a very interesting triangle. Okay. And the triangle was something like this. So this triangle was made by a very interesting, you know, I can say mechanism. Okay. So it starts with a one and ends with a one. Okay. So every row, let's say I call this as the 0th row. Okay. But there's, there's the reason why I'm calling it as a 0th row, not the first row. So all the rows that we are seeing over here, they are all starting and ending with a one. Okay. Of course, except the 0th row. And every intermediate numbers are actually obtained by adding the top two. Okay. This you have already seen before. Okay. So when you add these, this is what you're getting. So add one and three, you get a four, add three and three, you get a six, add three and one, you get a four, etc. Add this, you get this, add these two, you get this, add these two, you get this. Okay. So like that, this triangle can be, you know, this triangle goes on for ever and ever. Okay. Now, this guy, Abhay's Pascal, he, he found this in somewhere in 16th century. And later on, he moved into the field of, he became very religious. Okay. He gave up everything. And he went into the service of God. Okay. He, you know, gave up on his research work and, you know, findings in mathematics and he started doing Bhajan Kirtan. Okay. Not a bad thing to do, but later on, his work was seen by Newton. Okay. So Isaac Newton near about 17th century. And he realized that these numbers that Abhay's Pascal had left off, they actually find their place in these numbers. Okay. And he was the one who actually came up with the theorem. So, so Isaac Newton was the one who invented binomial theorem. And he realized from the Pascal triangle that these numbers that are written, they are coming from this listing in the Pascal triangle. So this one that you see is actually this one. Okay. And why did I call it as the zero throw? Because the power here is zero. That's why I call it a zero throw. Okay. The second one, one, one. You can see this is your one, one. Okay. The third one, one to one, one to one. Okay. Then one, three, three, one. See, one, three, three, one. So thanks to this Pascal triangle that now we will be able to fill up the other rows as well. So this will become, let me check. What was it? One, four, six, four, one. So this will become one, four, six, four, one. One, four, six, four, one. Okay. Similarly, for x plus a to the power five, I can use one, five, ten, ten, five, one. So let me write that down. One, five, ten, ten, five, one. Okay. And very surprisingly, these numbers that you see here, okay, these numbers that you see here in the Pascal triangle, they are called the binomial coefficient. Okay. So these numbers one, one, one, one, two, one, et cetera, they're actually named as the binomial coefficient because for the simple reason that they are coefficients of these expressions. Isn't it? So many times people ask me, sir, why do we call these numbers in the Pascal triangle as a binomial coefficient? Because these numbers basically serve as these numbers in the binomial expansion. Okay. And they're even more surprised when I, when I used to call NCR as a binomial coefficient. Okay. So fine. These numbers are seen in the binomial expansion. That is why you are calling them as binomial coefficient, right? But why are you calling NCR as a binomial coefficient as many people ask me? Why don't you just call it as N choose R? See, the reason is very simple. These numbers that you see is actually your zero C zero, one C zero, one C one, two C zero, two C one, two C two, three C zero, three C one, three C two, three C three, et cetera. So each of these listing in that Pascal triangle is actually an NCR term. It's actually an NCR term. And later, sir Isaac Newton actually made a sense out of it. Why it is, why NCR is appearing as a coefficient of this expansion, right? Which I'll discuss in some time, not to worry. So meanwhile, meanwhile, you can actually write these numbers as five C zero, five C one, five C two, five C three, et cetera. Right. So in light of this observation that you have, you can actually make a formula out of it. So he came up with a formula. Sir Isaac Newton came up with a formula. So he generalized it and he realized that if I am expanding a binomial expression having a whole number index, I could write it like this N C zero X to the power N A to the power zero, N C one X to the power N minus one A to the power one, N C two X to the power N minus two A square. Okay. I found this on the web for LC two. The moment it flashed, I came to know it got activated. Okay. Okay. Is it like any questions, any concerns? In short, he figured out that you can write this term as summation NCR X to the power N minus R A to the power R, R going from zero to N. So this particular expression that you see on your screen, this is what he called as the, or what we study as a binomial theorem for whole number index. Please note this down. Okay. And trust me, this result was only derived from observation, nothing else. There was no, I mean, of course you can prove it by using mathematical induction, but there was no derivation done to get it. You're just observing and then you realize that, oh, this is how this particular expansions work. Right. So this theorem came from our observation. However, I'll also show you a logical way to get to this expression in our next slide, but before that copy this down. Okay. So now another way to look at it is another perspective, I would say another perspective. So if you see X plus A to the power of N, if N is a whole number, you can write it like X plus A, X plus A, X plus A like this. X plus A times X plus A times X plus A. Okay. N number of times. Okay. N number of times. Now Newton, how did he read it? He read it as, he read it like this X or A. So he read, he read plus as or and multiplication as and. Okay. So he read it as X or A and X or A and X or A and X or A like that. Okay. Now when you multiply it, basically give you some terms. Correct. And now he said that every term that you will be getting will contain one of X or A from every factor. Okay, but not both. Okay. So either you can choose X or A from here, but not both. And every factor must contribute either an X or an A. Now, another easy way to understand this is think as if there is a country, okay, which has got N families. Okay. So this is a family. This is a family. This is a family. This is a family. So there are N families living in a country. Okay. Every family has a husband and wife. So X is a husband, A is a wife. Okay. Husband, wife, husband, wife. Okay. Now let us say this country wants to go for a war. Okay. So the government of this country gives a mandate or passes an order that every family must contribute only one person so that other person survives. Okay. So in how many ways can this army be prepared? So one way of preparing this army is you take all the X's from these families. So take all the husbands. Okay. So that is one way of making an army. Okay. So every family has contributed and every family has contributed the husband of the family. Okay. Another way to make that army is you take the husbands from N minus one families and take the wife from one family. Okay. That is also going to make, you know, N member and that is also going to make an army of N people. Okay. So you take N minus one husbands and one wife. That is another way to make it. But that one wife can come from any of these N families. So to choose that, you can have NC one way. So NC one way to choose the wife and the remaining ones will automatically contribute the husbands. Isn't it? So like that you can also have an army having two wives and N minus two husbands. And finally you can have, I mean, I'm just, yeah, finally you can have army of all the women's or all the wives. So this itself, you can generalize it as NC zero like this. Is it fine? Any questions? Any concerns? So this is another way of looking at the whole scenario. Alright. So whether this perspective or the perspective that you got from your pattern observation, few things are very, very clear, which I'll be taking up now in our analysis. So give me 10 minutes, then I will give you a break. So let me just finish off this analysis. The first thing that you would have observed that in your x plus a to the power N, where N is a whole number, the total number of terms, the total number of terms. Now, by the way, when I say total number of terms, I actually mean total number of dissimilar terms. Now, why I did not use the word dissimilar, you know, just when I said it for the first time, I just say total number of terms, I didn't use the word total number of dissimilar terms is because if you have x plus a and x and a both are dissimilar to each other, both are dissimilar type of term, what is the meaning of dissimilar? Not of the same nature. For example, one is a constant, other is a variable, nor one is a variable of one type, another is a variable of another type. So if you are having dissimilar terms, then you on your expansion also, you'll end up getting all terms dissimilar. If you have both the terms similar, you'll only get one term. For example, if I say two plus three to the power five, you get only one answer, five to the power five. So that's the reason why I'm saying these two terms are dissimilar, then only whatever I'm saying will make sense. So total number of dissimilar terms, when you are expanding x plus a to the power n is n plus one. Now, this can also be obtained by a permutation combination concept, which is basically nothing but the number of ways to distribute n identical objects between two distinct groups. As I told you when I was discussing the pattern that the index gets distributed between x and a, it's like, you know, you're distributing n chocolates between two people. So this is something which is very important and I'll be talking about it again when we do multinomial theorem. So first just note this down, total number of dissimilar terms, when you expand x plus a to the power n, when your x and a both are dissimilar to each other. So these two are dissimilar. Okay, is going to be n plus one. Is going to be n plus one. Is this fine? Any questions? Second thing I would like you to note down is your expression for the general term. Now, general term normally we prefer writing it as tr plus one-th term, where r could be any number from zero up till n. So as you know, if you want to express the first term, your r will be kept like a zero. That means you'll get t zero plus one, which is your first term. Okay, if you want to get the fifth term, r should be kept as four. If you want to get the last term, which is your n plus one-th term, r will be kept as n. Okay, so if you see the pattern, then you would realize that your r plus one-th term can actually be written like this. Okay, and you can check. See, if you want your first term, first term is n c zero, c zero, I'm putting a stress on zero, n c zero, x to the power n minus zero, a to the power zero, right? That is what has been obtained here. If you want third term, n c two, isn't it? This is your third term. This is your third term. So n c two, x to the power n minus two, a to the power two. So that is why this. Okay, so many people say, sir, this is not fair. Why didn't you write tr, my tr plus one? You could have written tr, isn't it? Why did you write tr plus one? See, if you write tr, you'll end up writing something very, very ugly looking, something like this. Oh, sorry, a. Now tell me, which is more easy to remember and which is more easy to write? Is this one easy to write or this one? First one or the second one? Obviously the first one. Second one looks so ugly. This looks so complicated. So this is more convenient to write. Okay, this is not convenient. Not convenient. And that's the reason why when we write our general term, we actually write tr plus one term, even though it sounds very weird because people will say, sir, in APGPHP used to write tr only. Why all of a sudden in binomial theorem, you started writing tr plus one? Because for the simple reason that tr plus one is easier to write as compared to tr. Yes, the only painful thing over here is whatever position you want, your r has to be one less than that position. So if you want the 70th term, you have to keep your r as 69. If you want your 100th term, you have to keep your r as 99. That is the only pain point there. Okay, so this is not preferred. This guy is not preferred, not preferred, and this is preferred. This guy is preferred. Okay, so going forward, when I write down the general term, I will use this expression for my general term. Is it fine? Any questions, any concerns? Now, third thing that we are going to discuss is your middle term, middle term or terms depends upon the situation. Now, frankly, speaking, this is a convention. Okay, now it is convention that when you are expanding x plus a to the power n, this is considered to be your first term. Why did I write i over there? Yeah, this is considered to be your first term by convention. This is second term, this is third term, this is your last term. Okay, so this is my convention that we follow. See, I mean, you have already done, let's say, you know, x plus a the whole square. If I write, if I ask five different people to expand it, one person will say x square plus a square plus 2x, one person will say a square plus x square plus 2x, one person will say x square plus 2x plus a square, five different people will give me five different ways to write it, isn't it? So is there any first term here, second term here, third term here is something like this sacrosanct? No, it is something that we have made as a convention that if there is an x and an a as the first term and the second term respectively, then this will be considered to be your first term. Let me just write it properly. This will be your first term, this will be your second term and this will be your third term. This is how we conventionally write it. Okay, so in light of this, this actually becomes your middle term then, right? Are you getting one more? So logically speaking, there is nothing like first term, second term, third term. It is not a sequence, right? So this is something that might arise in your mind. Who gives the authority to call that as a first term? There is no such authority. Just by convention, if you raise the entire power on the first term, that is your first term and if you raise your entire power on the last term, second term that is your last term and in between if that transition is happening, those are given corresponding numbers. Okay, so this is your first term, second term, third term. Okay, so in light of this convention, I'm discussing this part with you. So when you have x plus a to the power of n, the middle term actually depends on your n value. Okay, so there are two things that you will be talking here. If n is even, you tell me if n is even, which term will be the middle term or terms? So first of all, how many middle terms do we have and what will be the position of those terms? Okay, let's talk about n as even first. You can take some example, let's say x plus a whole square, we just noted. But by the way, I'm not going to write that expansion now. So you realize that t2 is your middle term, right? If you raise it to a power four, you will realize t3 is your middle term, correct? If you raise it to the power six, I'm not going to write them down. I'm just writing t1, t2, t3, t4. That's to tell you which term is the middle term. Then t4 will be the middle term. So how is this position? All of you tell me how is this position linked to this number? How is this three linked to this four? How is this four linked to the six? Can anybody tell me that? Right, Siddharth, there will be only one middle term that is very obvious from here. But what is the position of that middle term is what I'm interested in. There's one middle term. Okay, of course. But what is the position of that middle term? Is there any formula which you can use to link n value with the position? Does it work like that? Say two. Two into six minus two is right, Nikhil. So n by two plus one-eth term will be your middle term. Okay, please note that n by two plus one-eth term will be your middle term. See here, two by two plus one gives you two. Four by two plus one gives you three. Six by two plus one gives you four. Very nice. So this is your position of the middle term. So what is the value of that middle term? You'll say, sir, I already know this is my r value. So n, c, r, x to the power n minus r, a to the power r. So this is your position. Okay. And later on, we'll prove that this binomial coefficient is the greatest. So please note that the middle term has the greatest binomial coefficient. I've used the word binomial coefficient. We will talk about that also in some time. Okay. So remember, when n is even, there will be one middle term whose position will be n by two plus one-eth term. And that particular middle term is going to have the greatest of all binomial coefficients occurring in that expansion. Okay. So out of n, c, 0, n, c, 1, n, c, 2, r till n, c, n. If n is even, then n, c, n by two will be the greatest value. Now, why? We will prove this. We will prove this maybe in today's session. Let's see whether we reach to that extent. But if n is odd, if n is odd, then there will be two middle terms. Why? Because if you're raising it to an even, sorry, odd power, let's say I take odd, okay, one. So there are two middle terms. Both of them are middle terms only. Okay. By the way, I will write it as t1 plus t2. Okay. So these two are middle terms. If you raise it to the power of 3, then you have this. These two are middle terms. Okay. Yes or no? Raise it to a power of 5. Yeah. These two are middle terms. Okay. So can you relate these numbers that I'm showing with a circle? Can you relate these numbers with this index, with this white circle? Any relation between them? So it is obvious that they're two middle terms. No doubt about it. No debating about that. But how is that position of those middle terms linked to n? Can somebody tell me that? If you want, you can write a few more terms and check it out. Nikhil again, once again, where are you going Nikhil? So here you will have two middle terms. Okay. We had one middle in case of even, but we'll have two middle terms in case of odd. And what will be those terms? As you rightly said, Nikhil, it will be tn plus 1 by 2th term and tn plus 3 by 2th term. Okay. Right. Prism also is correct. So now if I have to write the value of that term, I'll better use my general term expansion. So I'll have to write it like this. How many people say, sir, why did you write it like this? Because remember, we had learned tr plus 1. So some r plus 1, I have to generate. So if you want to write this as an r plus 1, r will be n minus 1 by 2, obviously. So this will be ncn minus 1 by 2, x to the power n plus 1 by 2, a to the power n minus 1 by 2. And the other guy, you can write it as tn plus 1 by 2 plus 1, which is nothing but ncn plus 1 by 2, x to the power n minus 1 by 2, a to the power n plus 1 by 2. Okay. And just like we had it for our even terms, please note, ncn minus 1 by 2 and ncn plus 1 by 2, they would be both equal. And not only that, they would be the greatest of all the binomial coefficients occurring in that series. Okay. And this also I will prove officially, little later down the discussion. Okay. Is it fine? So these three, of course, we'll have many more things to be taken up in our problem solving. So these three things you should keep in mind. One is how many terms are there, which is very obvious, n plus 1. And second thing is what is the expression for the general term? This is what we follow. We normally do not follow this. So I will just score it off just to show you that we don't follow this. And finally, we discussed about what are them, you know, what is the middle term or middle terms, depending upon your n value, if n is even, only one middle term. And if n is odd, there will be two middle terms. And these are your positions, n by 2 plus 1 if n is even, and then plus 1 by 2 and n plus 3 by 2 when n is odd. Okay. This is something which I'll segregate over here. All right. All right. So time for a quick KitKat break. So 6.17 is what is the time right now? We'll meet at 6.32pm. Okay. On the other side of the break, we'll see some problems based on whatever we have done so far. Okay. See you on the other side of the break. So let's start with some basic questions on whatever we have covered so far. Let me begin with this question. Expand this by binomial theorem. Okay. Now it's a very simple question, but I just wanted to see whether everybody has understood that basic theorem that we had discussed in the initial part of our class. Now don't simplify any term. Just write them down. So open your notebook and just write down the six terms. Why six terms? Because the power here or exponent here is 5, right? So write down the six terms that you get from the expansion of this term. Okay. No need to expand anything. Just to give you a hands-on practice and let me know once you're done so that we can immediately discuss it out. Just a simple done on the chat box is sufficient enough. Is it done? Now please check your answer. The terms would be 5C0 to A to the power of 5 minus 3 by B to the power 0. 5C1 to A to the power 4 minus 3B to the power 1. 5C2 to A to the power 3 minus 3 by B to the power 2. 5C3 to A to the power of 2 minus 3 by B to the power of 3. 5C4 to A to the power of 1 minus 3 by B to the power of 4. And last is 5C5 to A to the power 0, negative 3 by b to the power of 5. So one important thing that came out from this exercise is that this sign that you see over here, minus sign, it is actually considered to be a part of the second term. So this term should not be just treated like 3 by b, it should be treated with a minus sign. That is why if you see in all these expansions, I have written minus 3 by b as the second term. This is a mistake which many people do initially. Of course, not in the long term, they do not do this mistake. But initially, people forget to include this sign. So how many of you actually forgot to include that sign? From here you realize that the terms will be alternately positive negative. As you can see, this will be a positive term. Second will be a negative. Third will be positive. Fourth will be negative. So this leads to an alternative positive negative term. Alternative, positive, negative. I mean, I would say there is an alternative positive negative sign in between. I should not say positive negative term, but there are alternately positive negative signs in between. So whenever you realize that there is an alternation of sign positive negative, it definitely implies that in the binomial expression, there must be a negative in between. Is it fine? Any questions? Any concerns? Okay, so we'll now see a very advanced version of this type. Okay, let's take this question. Find the sum of the series, summation from 0 to n minus 1 to the power r ncr, half to the power r, 3 to the power r, 2 to the power 2r, 7 to the power r, 2 to the power 3r, 15 to the power r, 2 to the power 4r, up to m terms, up to m terms. It's just based on your understanding of the binomial theorem, nothing else. It is one of the basic questions or basic concepts it is based out of. Yes, this is ncr. If you want, I can write it in a proper way. This is ncr. Okay, anybody has any idea? Okay, let's discuss it out. See, if you see this term, let me write this, let me call this as, by the way, sm. I'll choose a green color here, sm. Yeah, sm because you're summing up till m terms. Now, each of these terms of sm, I'm going to write it separately like this. Summation r equal to 0 to n minus 1 to the power r ncr and I'm going to separate this term. Okay, second term likewise will be summation minus 1 to the power r ncr r equal to 0 to n. Second term is 3 to the power r by 2 to the power 2r. So the whole thing, I can write it as 3 by 4 to the power r. Do you all agree with me? Can this be written like this? Obviously. Next term, likewise, I'm going to write it like this, 7 by 8 to the power r. So this goes on the way till m terms. Now, I would request everybody to have a deeper look into each one of these terms. Look at this term. If you look at this term and if you start writing it, if you start expanding it, the first term will be nc0, half to the power 0. Second term will be minus nc1, half to the power 1. Third term will be nc2, half to the power 2. Then minus nc3, half to the power 3. And this goes up till minus 1 to the power n, ncn, half to the power n. Can you relate this to a binomial expansion? If yes, whose binomial expansion is this? Are you able to identify? No? Okay. Then let me add one more thing here. Let me put n1 to the power n, 1 to the power n minus 1, 1 to the power n minus 2. Right. Absolutely right, Nikhil. So this is actually 1 minus half raised to the power n. Do you all agree with me on that or not? Agreed. Likewise, can I say this term is nothing but 1 minus 3 by 4 to the power n. This term is 1 minus 7 by 8 to the power of n. In short, your series contains half to the power n, 1 by 4 to the power n, 1 by 8 to the power n, 1 by 16 to the power n, and so on up till n terms, up till m terms. Correct. Agreed. Agreed or not? Agreed or not? Agreed? Okay. So this is nothing but it's like a geometric series. It's like a geometric series. That means it is the sum of a geometric progression. Whose first term is 1 by 2 to the power n? Common ratios also 1 by 2 to the power n check. If you multiply this with 1 by 2 to the power n, you get the second term. If you multiply this with 1 by 2 to the power n, you get the third term. And total terms, total number of terms is going to be m because there are m such terms. So can you not apply geometric series formula to sum this up? Right. So what is the geometric series formula for m terms? Tell me, tell me, tell me a to the power, sorry, a into r to the power, total number of terms minus 1 by r minus 1. So in light of this, I can write it as 1 by 2 to the power n. Common ratio is this whole a to the power m minus 1. Okay. By this, on simplification, this will give you 1 minus 2 to the power mn divided by 2 to the power mn 1 minus 2 to the power n, which you can also state it like this. Both are same things. This is going to be your answer to this particular question. Is it fine? Any questions? Any questions? Any concerns? Okay. So it was all about identification, whether you are able to identify that each of these terms is actually a binomial expansion in itself. So you are actually summing up binomial expansions. Right. So each of these terms was actually a binomial expression in itself. And you're finding the sum of those binomial expressions. Very good question. Okay. All right. Let's move on to the similar type of questions. In fact, okay, let's say, okay, let's take some question on middle terms. Find the middle term in this expansion. Find the middle term in this expansion. Middle term or terms that depends upon the expression. You can just say it done also. No need to actually write your response. Okay. Just say it done if you're done. Great. So here we know that n is odd number. Right. So this is an odd number. So we had already discussed that there will be two middle terms. Right. So two middle terms will come out. Two middle terms will come out. Right. What will be those two middle terms as we already discussed? Tn plus 1 by 2 and Tn plus 3 by 2. Okay. Which is nothing but T5 and T6. Now, how do you find T5? T5 is nothing but T4 plus 1. See, always while you're finding out the general term, you should always write it as R plus 1. So if this T5, your R will become automatically 4. So this will be your R. Okay. So now use the formula nCr first term to the power n minus R, which is this. Second term, by the way, second term will include the negative sign within it. Don't exclude the negative sign. Okay. So in short, it will become 9C4 3x to the power 5 minus x cube by 6 to the power 4. Okay. Similarly, this will become T5 plus 1, which is nothing but 9C5 3x to the power 4 and minus x cube by 6 to the power of 5. Now don't need to expand it, but these two will be your middle terms. That's it. Okay. Don't waste time simplifying it. It's fine even if you leave it till here. Is it fine? And you can check yourself that nC5 and nC4 will be equal and there will be the greatest of all binomial coefficients. There's something which I would like to address over here. What is the difference between term binomial coefficient and coefficient? Okay. Let's try to understand this. Okay. Let me take a simple example. If somebody says 2 plus 3x, okay, expand this, 2 plus 3x square, let's expand this. So if you use our binomial expansion, we'll end up getting something like this. I've just taken a simplistic case so that we don't waste too much time. Now let's see if I ask you what is the second term. This whole thing will be your second term. Okay. So this whole thing that is 2 to the power 1, 3 to the power 1x, this whole term will be your second term. Getting the point. Just give me a second. I think the screen would have got unshared. Just give me a second. Yeah. When power goes off, the screen gets unshared. Yeah. Now what is the binomial coefficient? Binomial coefficient of, let's say, second term is your, this part, this is your binomial coefficient. Only the 2c1, that is the binomial coefficient. Okay. So normally they will say what is the binomial coefficient of fifth term, third term, sixth term, whatever, whatever term. There you just have to mention the ncr which is sitting in that term. If somebody says what is the coefficient, now normally the word coefficient is something which is, the word coefficient means something which is sitting along with something. So we say coefficient of some expression, like coefficient of x, coefficient of x square. So if the word coefficient of something is not mentioned, only coefficient is mentioned, then don't get panicked. It automatically means coefficient of the variable. Okay. Whatever is the variable involved, the number associated along with that variable is your coefficient of that variable. So in our case, if they say coefficient or coefficient of x that will comprise 2c1, 2 to the power 1, 3 to the power 1. So this whole thing will be coefficient. Are you getting my point? So it is very important to distinguish between at least the last two which I have written. Term is obvious. Many people don't get confused in that term. But what is the difference between binomial coefficient and coefficient should be very clear. Now, I am not stating that coefficient cannot be binomial coefficient. So if let's say there was no constant involved other than the binomial coefficient, then that itself will be your coefficient. Okay. So your binomial coefficient can also be your coefficient provided that there is no other constant involved with it. Here you have a 2n3 also sitting here. Right. So your coefficient is different from your binomial coefficient. But let's say if you had only 1 plus x to the power of 2. Okay. Then in this case, just give me a second. Some children are screaming like, yes, if you have something like this, then you will only get 1, 2c0, 1 to the power 2, x to the power 0, 2c1. In fact, we don't write it actually like 1 to the power 1. So 2c1, x, then you'll have 2c2, x squared. So if you somebody ask you what is the binomial coefficient in the second term? So this is your term number two. So here your binomial coefficient will be 2c1 and that will be also your coefficient. That means coefficient of the variable. Variable here is your x. Okay. So I'm not trying to say that binomial coefficient and coefficient will always be different. There will be occasions when your coefficient itself is your binomial coefficient. So that will depend on question to question. Okay. So this is very important. Don't make a mistake about it. Okay. So are you clear about the difference between what is the difference between these three terms? I hope nobody's going to make mistake based on these simple concepts. Okay. See, why I'm telling you this is because in your past, your seniors have made mistakes. Okay. They say, oh sir, they were asking coefficient. I found out binomial coefficient. Gone. Right. Four marks gone. Okay. Anyway, so now let's make some problems based on general terms. So let's start with the simple question. Find the seventh term in this expansion. Now again, as I told you, by convention, we call, okay, so I mean, starting with 4x as the first term and minus 2 by root x as the second term, whatever we get, the first term is that 13c0, 4x to the power 13 minus half root x to the power 0. That will be your first term. And accordingly, when you keep writing it, they are called second term, third term, fourth term and so on. Okay. That is by convention that we follow like that. Anyways, so what is the seventh term of this particular expansion? Just see it done on the chat box if you're done. No need to give me a response. I don't want anybody to sit and type that out. Done very good. Okay. So this is just based on your understanding of the general formula. So tr plus one, we already know it's ncr, the first term raised to the power n minus r, second term. Second term means second of that expression raised to the power. Right. So here, just understand that t7 means t6 plus 1. Okay. So as per your formula, which you have written over here, it will be 13c6. That first expression is 4x. n minus r is 7. Second expression is minus 1 by 2 root x raised to the power of 6. Okay. And again, it is up to you to simplify it, but I'll just do a basic simplification. If I'm not mistaken, this is 2 to the power 14 and you have 2 to the power 6 here. This is x to the power 7. This is x to the power 3. So this will boil down to 13c6, 2 to the power 8, x to the power 4. So this will be your seventh term of that expansion. Fine. Any questions? Easy. So these are all basic questions which I'm taking up on the topic. Should we move on to the next one? Okay. Let's take this one. Find the fourth term from the end. Find the fourth term from the end of this expansion. Again, write a done if you're done. No need to type out your response. Excellent. So many of you have done it. Now see, when you're finding the fourth term, especially from the end, this is very, very important. Many people, what they do, they literally start counting from the end. Actually don't need to do that. When somebody says fourth from the end, you just reverse the position of the first and the second expressions. And find from here, from here you just find out your fourth term, the usual way of finding the r plus 1 term. So you don't have to back count it. Just reverse the position of the first and the second expression. And just find out your normal r plus 1 term, like we have learned in our analysis. So t4 will be t3 plus 1. So it will be 7c3 minus 2 by x square to the power of 4 x cube by 2 to the power of 3. Now very important to note here is that the minus sign is actually shifted along with the term. Okay. As you can see, I've just not swapped the position around the minus sign. This minus sign is actually considered to be the part of the second term. So it is red like this. Getting one more. So when you're swapping their positions, the minus sign will be accompanying that 2 by x square everywhere. Getting one more. Of course, again, if you simplify it, if I'm not mistaken, this will give you 7c3 2 and 2x, which is actually 70x. Is this what you people are getting? Those who said done, 70x? Is it fine? Okay. Let's take your questions based on finding the coefficients of a certain power of x. Okay. So let's take this question. Find the coefficient of x to the power 8 in the expansion of this term. Okay. Now in this type of a question, you don't have to actually write all the terms. And you should not be actually doing it because writing all the terms means writing 11 terms. And then you're scanning each of these 11 terms to see which term is generating x to the power 8. I would not suggest you to take that approach because that is going to be too lengthy. Okay. And in an examination scenario, you are not going to get that liberty to spend that much of time. And imagine what will be the price if their power was, if that index was more than 10, let's say it was like 50 or something. Okay. Will you sit and write all the terms? So here the approach is we assume that let r plus 1th term contain x to the power 8. Okay. So I don't know which term is going to give you x to the power 8 finally, but I'm just taking a guess or just taking an assumption initially that let r plus 1th term is the one where x to the power 8 actually appears. So as per our expansion of the general term, this will be 10 cr, this to the power 10 minus r minus 1x to the power r. So here separate out the constants from each other and separate out powers of x. Okay. I hope everybody has understood till this extent. So 10 cr. Okay. I've kept it separate. Minus 1 to the power r will come from here that I've kept it separate. And if you consolidate your powers of excess using your exponent laws, you end up getting x to the power 20 minus, oh, oh, 2r. No worry. 20 minus 2r minus r coming from this last term. Now, if you want that this term should contain x to the power 8, which means this term should be 8. That means 20 minus r, 20 minus 3r should be 8, which means 3r is equal to 12. That means your r value is equal to 4. Now, they're asking you for the coefficient. They're asking you for the coefficient. So coefficient is this way. Please note that everything along with the variable that is called coefficient, everything which is sitting along with it. I mean sitting with it. Okay. That is called the coefficient. So your coefficient is 10 cr minus 1 to the power r. r is 4 here. And this answer is 210. Okay. 10c4 is 210. I hope everybody remembers that value. Now, one important thing to note over here. If your r doesn't come out to be a whole number, it actually means that the particular power of x that you are looking for does not appear in the expansion. Okay. So let's say while solving it, here we were lucky to get a whole number. But let's say you got 4.5. Let's say you got a minus 2. It just indicates that that particular power of x is not there in that expansion. Okay. For such case, you will write the coefficient to be 0. So please note this down. Okay. You will not say coefficient does not exist. You will say coefficient is 0. So if a certain term does not appear, we basically take its coefficient to be 0. Okay. There was one question I think it was asked in one of the regional interest exams where they gave this option as coefficient does not exist. Coefficient is 0. And most of the people went for coefficient does not exist. No. If some term is not there, its coefficient is 0. That's why the term is not there. Is it fine? Any questions? Any concerns? Is this approach understood by everybody? Okay. So now we'll take a similar question. Let's take this one. Find the term independent of x. Find the term independent of x. Independent of x means it will just be a constant. There will be no x there. Okay. So it will be a straightforward constant value. Okay. So which term in this expansion will be just a constant is what the question is asking you. And again, no need to type your response. Just say a done that is more than enough for me. Okay. Can you also give me the value? Can you give me the value as well? If at all, I mean, it's not too big to write. See, you don't have calculators in your exam. I hope everybody is aware of it, neither in your CVC exam nor in your cognitive exams. Okay. So if the number is too big, you can leave it like that. Okay. And if you can find it out by using your, you know, normal calculations, please find the value as well. In case you're able to find out the value, if it is not too big for you to find it out, please find it out. Okay. Prisham is also done. Prisham, I hope you have not just stopped at finding the position. You should also find out its value. If at all it is, you know, calculatable. Okay. Should we discuss it? So just like the previous question, we'll say, let R plus 1 at term be independent of X. Okay. So let's say this is independent of X. In the same way that we used in the previous problem, we will write down the R plus 1 at term. In this case, it will be 9 CR, 3 by 2 X square to the power of 9 minus R minus 1 by 3 X to the power of R. Okay. Let's consolidate, let's consolidate all the constants together. Okay. So that is the first thing I'll do. And then we consolidate all the powers of X. If I'm not mistaken, these will be all powers of X. Correct me if I'm wrong. Yeah. Okay. Now you're claiming that it is independent of X. Independent of X means the index or the exponent or the power should be zero, which means 18 minus 3R is zero. That means R is 6. Thankfully, it's a whole number. That means there is a term. Okay. There is a term which is independent of X. So what is that term? Let's find it out. So the term independent of X is your seventh term and that term is 9C6, 3 by 2 to the power of 3. So let me write it like this. Okay. Minus 1 by 3 to the power of 6. That's just 3 to the power 6. Okay. 9C6 is 9 into 8 into 7 by 3 factorial, which is 6 into, into, into, into, into. Why did I write a 3? It should be 6, no? 6, 6. Yes. So this will be 3 to the power 3, which is 27. And we'll also have an 8. So 8, 8 gone. 7 by 18. 7 by 18, dear students. Is this what you are getting? Tell me. 7 by 18. Those who said yes or those who said done, 7 by 18. Is this what you're getting? Okay. All right. So with this, we'll now move on to another type of question. In fact, a very interesting one. Before that, if you want to ask me something or copy something down from here, please do so. Okay. So let's move on to this question. Is it? Yeah. Find the number of irrational terms in this expansion. Find the number of irrational terms in this expansion. So as you can see, the power is 100. Okay. So when you expand it, you'll get almost, almost what? Exactly. 101 terms. So out of those 101 terms, how many terms do you think are actually irrational in nature? Now this is what the competitive exams want you to solve. The previous ones were all basic copy book questions. So these were the ones, these are the ones where you have to think of it. So let me give you some time. Once you get the answer, please give it, give your response on the chat box. Yes. Any idea how to do this? Just a minute. Okay. Sure. Sure, Nikhil. I'll give you some time. One minute, one and a half minutes. Whatever is necessary. Take it. No issues. This is 5 to the power of 18, 2 to the power of 1 6th to the power of 100. So they're asking you how many of these expressions will be, or how many terms of this expansion will be? Thirds. Excellent. Very good, Nikhil. Awesome. So Nikhil has solved this question. Very good. Anybody else? Okay. Never mind. Let's, let's focus on any general term. Okay. So if you take any general term of this expression, you will have something like this. Okay, Siddharth, Karthik. Good try. But unfortunately, that's not the right answer. We'll check it out. Okay. So if you write it down like this, you'll have something like this coming up. Okay. Now let us focus on what will make this term rational, first of all. Let's say if I want this to be rational. Okay. What can you comment about these powers that you have on 2 and 5? Let me hear it out from you. Should they be, I mean, any value or there should be some specific values assigned to them? Let's say, can I have this as 2 by 3? Or can I have this as let's say 5 by 8, something like that? If you want the entire term to be rational, q, q means rational, what can you comment about these two expressions? Can I say they should be integers? Any integer? Positive, negative, whatever. It should be integer, right? Because if that term is not an integer, there would be a third hidden. And let me tell you, because this 2 and 5 are, you know, relatively prime, that means they will not have even some common factors to make that 2 served as a rational number. Okay. So 2 and 5 being relatively prime to each other, they will not have any kind of a common serve between them to make them, to basically remove that third thing from there. For example, let's say both of them contributes root 2, root 2. So root 2, root 2 may become 2, which is a rational number, but even that is not going to happen. So the only way that you are going to ensure that this term remains a rational number is by ensuring the powers of 5 and 2 remain integers, right? So now we have to figure out that how many rs exist between 0 to 100? Please note that your r can go from 0 to 100 for which both of them are integers. Okay. So this also is an integer and this also is an integer. How many such r exist? Okay. Now let us figure out the, let's focus on our second expression. So between 0 to 100, you can have the following rs for which r by 6 is an integer, till 96. Okay. Now which out of them is also making 100 minus r by 8 as an integer? Let's figure that out. So what I've done first, I figured out when is this integer first and then for those values of r, which will make r by 6 integer, I'm trying to again select which one is making this guy also as an integer. So finally, whichever satisfies 100 minus r by 8 as an integer from these values, they will only be taken in our consideration. Okay. So let's start with 0. 100 minus 0 by 8, no, not an integer. 6, 94 by 8, not an integer. 12, 88 by 8, yes, it's an integer. So 12 is fine. Okay. 18, no, 72 by 8, sorry, 82 by 8 is not an integer. Right. 24, no, 76 by 8 is not an integer. Then the next number would have been, I think 24 is over, yeah, 24 is over. Next number will be, you let me write it down, 24, 30, 36, 30, 30, no, not an integer, 36. Yes, 36 will give me an integer because 36 will make the numerator as 64. Okay. So let me just write it down, slightly more. Okay. By the way, I will not write all of them. There's a pattern by which you can actually identify. Now the next number that you will get will be at a gap of 24. Okay. So the next number that you will be putting is 60. So 100 minus 60, that is also going to be, okay. Next will be 84, 100 minus 84 is 16, 16 by 8 is an integer. Okay. So these are the only four numbers that only four value of r's for which you will end up getting this quantity as a rational number. So that means only t13, t37, t61 and t85 are rational. Rest all are irrational. So how many will be rational? Which means 101 minus four because only four of them are rational. So 97 irrational terms will be there. Is it fine? This is the type of question that you are expected to deal with in your comparative exams. Is it fine? Any questions you have in the solution of this particular problem, please definitely ask. Okay. Any questions? Nobody? Are you happy with the solutions? Okay, fine. So let's take a similar one, a very similar one. So exactly the same approach has to be used in the next question, which I'm going to give you. So if you have done with the copying part, I'm going to the next slide. Okay. Let's take this question. Yeah. Find the sum of all rational terms in this expansion. Find the sum of all rational terms in this expansion. This question I'm expecting most of you to get the answer. Okay. I should be seeing my chat box flooded with answers. Exactly the same approach. Only differences, they're asking you for some of all the rational terms. So whatever rational terms you get, you have to add that. Yeah, anybody? I thought people will be done by now. Okay. I can give one more minute. Okay, Tejaswini. Sum of all rational terms in this expansion. So whatever rational terms you're getting, you have to add them and get me the give me the sum. See, let's let's again write down the R plus one term. R plus one term is 15Cr, 3 to the power one-fifth to the power of 15 minus R and 2 to the power one-third to the power of R, which means nothing but 15Cr, 3 to the power 15 minus R by 5, 2 to the power of R by 3. Oh, very good Prisham. Very good. Excellent. Anybody else? Okay. Now, again, if you want this guy to be rational, you must ensure that 15 minus R by 3 is an integer and so should be R by 3. So both of these guys must be integer each. And your R can only go from 0 to 15 because your highest power is 15, the power here is 15. So R can start all the way from 0 and go to 15. So between 0 to 15, how many whole numbers meet this criteria that 15 minus R by 3 is also an integer and R by 3 is also an integer. So first we'll focus out how many makes R by 3 as an integer. So 0, 3, 6, 9, 12, 15. Now out of this, how many will make this as an integer? 0, will it make it as an integer? Is 15 minus 0 by 3 integer? Yes. Is 3 wicking an integer? No. 3 will make 12 by, sorry, this was 5, I believe. So 0 is going to make an integer, but 3 is not going to make an integer. 6 also? No, 9 by 5, not an integer. 9 will give you 6 by 5, not an integer. 12, 3 by 5, not an integer. 15, 0, that is an integer. So these are the only two terms. These are the only two terms and this is actually your first term and the last term. These are only rational. These are only rational. Okay, so let's find out T1. T1 will be nothing but 15 C0. This whole thing to the power of 15 which will make it 3 to the power 3 which is nothing but 27. Okay, and T16 which is nothing but 15 C15, 2 to the power 5. So that is nothing but 32. So the sum of T1 and T16 is nothing but 27 plus 32 which happens to be 59. So those who said 59 and only one person said that, Prisham absolutely right. So 59 is the sum of all the rational terms. Is it fine? Any question? Was this problem solved in a very different way visibly the previous one? No, it was almost the same procedure involved. Is it fine? So last five minutes, we'll take one question and we'll call it a day. By the way, this chapter is only 30% covered. Less 70% I would try to cover up in our next class. So meanwhile, we'll take a small question to conclude the day. Okay, let's take this question. Okay, just a very interesting fact. So this fact has been stated that 101 to the power 50 will have more value than 100 to the power 50 plus 99 to the power 50. Add a doubt in the previous one. Okay, let's go to the previous one. C2 has some doubt. Yes, tell me what is the doubt? Why can it be only be 0 and 15? Because 0 and 15 are the only values which will make both of them as integers. Check it out. 0, 5, integer, every integer, 15, baby, every integer, every integer. Rest no other values will make both of them integers other than 0 and 15. Out of these, out of these only these are the ones which made the criteria. This only makes this guy as integer. So this is ruled out. This is ruled out. This is ruled out. This is ruled out only 0 and 15. Same thing I did in the previous question also. Okay, so this is a very interesting fact that even when you combine 100 to the power 50 and 99 to the power 50, it is unable to beat 101 to the power 50. Okay, can you prove this by using binomial theorem? Can this be proved using binomial theorem? Of course, you can't sit and calculate their values because they're two bigger numbers. Okay, so without actually calculating their exact values, how can you establish this inequality between them? Okay, so in the interest of time, I'll be solving this problem because there's just two minutes left. I don't want it to extend beyond. Okay, okay, we'll check this out. See 101 to the power 50, can I write it as 100 plus 1 to the power 50? Okay, and when you write this expansion, it becomes 100 to the power 50, 50 C 100 to the power 49, 50 C 200 to the power 48, till the last term will be I think 50, let me write the pen ultimate term also 50 C 49 100 and last is 50 C 50, which is one actually. Okay, now we also have a 99 to the power 50 also sitting, can I write that as 100 minus 1 to the power 50? So basically, I'm trying to use 101 to the power 50, 99 to the power 50 and trying to write it in terms of 100 to the power 50. Okay, so in this case, you'll realize that there would be an alternating plus positive negative signs. Okay, this will be minus 50 C 49 100 and last will be again 50 C 50. Now, these two expressions, let us subtract them. Let us subtract these two expressions. So when you subtract them, you get 101 to the power 50 minus 99 to the power 50 and while subtracting, you realize that these terms will start cancelling off. Okay, and you'll end up getting twice of 50 C 100 to the power 49. So only these alternative terms is going to survive. Correct. I think the last term that is going to survive is 49, sorry, 50 C 49 into 100. Okay, now let us open this particular bracket to a certain extent. So let me just multiply two with this guy. So two with this guy will give me all of you see here. This is two. This is 50. So two into 50 into 10 to the power into 100 to the power 49. Won't that become 100 to the power 50 itself? And other terms, I'm not going to disturb them. I'm just going to write as they are. Okay, there's no point writing them down. Okay. Now, my purpose is solved here itself. So here you realize that 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is this fellow common sense is that this is a positive term. Correct. Which means 101 to the power 50 minus 99 to the power 50 minus 100 to the power 50 is positive term, which means 101 to the power 50 is greater than 100 to the power 50 plus 99 to the power 50. So even when you combine the two, it is not able to supersede. It is not even able to equal to it. By the way, it is actually much more than that because as you can see this term is going to be huge. So 101 to the power 50 is much, much, much, much more bigger than the combined sum of these two. Okay. Okay. So dear all, we are going to stop here. This chapter has just started by the way. Next class when we meet, we are going to talk about the concept of numerically greatest term, numerically greatest coefficient. We are going to talk about applications of binomial theorem. And of course, binomial coefficient series, binomial coefficient for any, any index. And finally, we're going to talk about multinomial theorem. So let's see how much we are able to conclude in our next session that we have. Thank you so much. Bye bye. Take care. Stay safe. And please get yourself vaccinated ASAP. Okay. Signing off. Good night. Thank you.