 Hello, I'm Zor. Welcome to Unizor Education. Today's topic is another problem in the category of mathematical induction. As the previous one, it will not be simple or trivial, but still not very difficult neither. It's a very nice exercise and it's a famous binomial formula by Newton. First, a couple of preliminary notes, if you wish. The most important purpose of this particular problem is to prove the formula which basically expands this particular expression. Before starting anything about this, let me just exemplify it in a couple of cases when n is a very small number like 2 or 3. You probably know this, but anyway I'll just derive a couple of formulas for n equals 2 and 3 and we will see some kind of pattern here. For n equals 2, a plus b square is equal to a plus b times a plus b, which is a plus b first multiplied by a. We will get a square plus a b and then we multiply a plus b by b. We're using a distributive law of multiplication. a plus b times b will be a b plus b square, which is equal to a square plus 2 a b plus b square. So that's the final formula. Okay, fine. Let's try something else. Let's try for n equals 3. a plus b to the third degree would be, I will use a plus b square, which I already know how to decompose n times a plus b. So this is a third degree. This is 2 and 1 and the powers are edging if you're multiplying one to another. So this would be a square plus, I'll just substitute this expression, okay, equals 2. First we multiply this by a and we will get a cube plus 2 a square b plus a b square. I'm trying to look like a before letter b in all the cases. Now we multiply it by b with the a square b plus 2 a b square plus b cube equals a cube and again here I will try to use the power of a as the dominant power which will be decreasing from 3 to whatever is necessary. Now a square b and another a square b. So it's 2 a square b and a square b. It will be 3 a square b. Now next will be a to the first degree a b square and 2 a b square. So it's 3 a b square and b cube. What I would like you to do is look very carefully to this and this. What's interesting is, you see this is the second degree, the power of 2. This is the power of 2. This is 1 and 1, power of 1, power of 1, which summarized together we will give you 2 as well. And this is the power of 2. So basically every member, every element of this sum has a combined sum of its powers equal to exactly the power of the original binomial. Now same thing here. You see here original sum a plus b was powers of the third degree. Now this is 3. This is 2 plus 1, 3 again. This is a to the first degree and b to the second. Sum is 3 again and this is 3 again. So as you see there is some kind of a pattern that if you will get a plus b let's say to the end of the degree. Most likely you will get a sum of elements of a in sum degree and b in sum degree. By the way I can always put here times b to the power of 0 because you know that every element, every number being brought to the power of 0 will give you 1. And same thing here. Instead of just b cube I can put a to the power of 0 times b. So every member is a to some power and b to some power but sum of 2 powers is equal always to the original power of the binomial. And then there are obviously certain coefficients which we don't know anything about yet. We will by the end of this lecture. Same thing here. I can always say that this is b to the 0 and this is a to the 0. So it's always a to something and b to something but sum of these two powers is always equal to original power of the binomial. All right this is just an observation. Now I will write down this particular formula and then I will prove it using the method of mathematical induction. Basically that's the goal. And by the way this formula in this particular case will be as the beginning of the base for mathematical induction when we basically check the formula for n equals to let's say two in this particular case. So let me write down the formula and then we will start the proof. So for n is equal to 2 it will be a b 0 plus 2 a first b first plus a 0 b square. Here's my formula for any n. And then I will explain what I mean. First of all what I have written here is basically what I said before that a plus b to n's degree is a sum of certain elements. Each element has certain coefficient and in this case the whole coefficient is basically expressed as this type of symbol. Let's just consider this to be a symbol. Then we'll talk about what exactly the meaning of this symbol is. But anyway in decreasing powers of a you see I put a to the n's a to the n minus 1 a to the first and the last one a to the power of 0 which is 1 actually. And b will be increasing b to the power of 0, b to the first, b to the etc, b to the n minus 1 and b to the n. And the sum of these powers as you see n plus 0, n minus 1 plus 1, 1 plus n minus 1 and 0 plus n. It's all the same, it's n. So basically I have expressed in this formula which is called by the way the binomial formula proved by Neutron, I mean Cereisic Neutron in whatever 16 or 17th century or whatever time we go on. So this formula represents the a plus b to the n's degree. Now what I did not talk about yet is what these coefficients represent. Now in this particular case for n is equal to 2. The first coefficient is 1, then 2 and then 1 again. Now what are these in general case? Well here is the definition. n factorial divided by k factorial and n minus k factorial. Now what factorial is just in case somebody does it now, n factorial is 1 times 2 times 3 etc times n. So multiplication of all the numbers from 1 to n. Alright we consider this to be an obvious fact. Now I have just written the formula. I can't really talk about what exactly is the proof of this formula, what's the approach etc etc. Now the approach is mathematical induction. We will prove this formula using mathematical induction. It's much easier by the way because if you know the formula, it's easier to prove it using the method of mathematical induction than to invent it from scratch. So you are in much better position than Sarai's Neutron 400 years ago or whatever. Now being in a much better position still doesn't mean it's an easy problem. You still have to really do the proof. And right now I will start the proof of this theorem using the method of mathematical induction. Before that I would like you to think about proving it yourself. You can just press the pause button and do it. And as a hint I can just tell you that basically what I will do I will use this formula for n as a given because we are assuming that the formula is true for some number. And then for the next number for n plus 1 or for n equals k plus 1 whatever. We will try to use the formula for n elements for the power of n. We will try to use it to prove the next one for n plus 1. So right now it's a difficult time to press the pause button. And I will start this particular proof. Okay, so what we have to do, we have to make a very important step from n to n plus 1. So if we consider that this formula is true for n, how does it look for n plus 1? Let's just first write it down. Well, it will be a plus b to the power of n plus 1. It should be equal to, that's what we have to prove, n plus 1 to 0, the coefficient. I'm just changing from n to n plus 1, a n plus 1 times b to the 0 plus n plus 1, a, now instead of n I'm using n plus 1, so it will be this. Some of these two powers now should be always equal to n plus 1. And the last two members will be n plus 1 to n, a 1 b n plus the last one will be n plus 1, n plus 1, a 0 b n plus 1. That's what we are supposed to prove. Now, how can we do it? Well, as usual, we will use this formula as given and how to get to a plus b to the n plus 1. Well, we'll get a plus b to the nth degree and multiply it by a plus b. So we will multiply the whole expression by a plus b and we will try to prove that this will be equal to this expression. Sounds easy, right? Okay, let's start. I will wipe it away because it takes a very important real to my board and we will start from a plus b to the n plus 1 equals to a plus b to the n times a plus b. Okay, great. For a plus b to the power of n, we will use this expression which we consider to be as given and then we have to multiply it by a plus b. Now, what does it mean to multiply by a plus b? Obviously, we will multiply it by a and then add it to the same sum multiplied by b. So, when we multiply it by a, a to the nth will be converted to a to the n plus 1, a to the a minus 1 will be converted to a to the nth, etc. And when we will multiply it by b, we will increase the power of b correspondingly. And so the first sum multiplied by a and the second sum multiplied by b will contain also a to some degree and b to some degree, but then I will write it as two lines, one under another and I will try to align them in such a way that the elements with the same powers of a and b will be one under another. And that's how we will do it. Let's start with multiplication by a. So, it will be n to 0, a n plus 1, b 0 plus n 1, a to the n, b to the first degree, a n minus 1, minus 1, let's be accurate, a square b n minus 1, right? We are multiplying by a, so every member has a power of a increased by 1 and the last one will be n 0, a to the first degree and b to the nth. Okay, that's what we receive if we multiply it by a, this expression by a. Now we have to multiply it by b and every member will be multiplied by b, but I would like to write it in such a way that elements with the same powers will be under these each other. So this m to the nth and multiplied by b will be b to the first, so I will put it here, shift it, n 0, a to the nth, b to the first, plus 1. Then let me go to the last member, so that's two members. This one will be a to the 0, b to the n plus 1 because multiplied by b. This one will be a to the first degree and b to the nth degree. So these are two last members. Now this will obviously should go under this one, sorry, n minus 1, a to the first, b to the nth degree. And this one will be an extra member which I will put as n to the n, a 0, b, n plus 1. You just write n plus 1, it's not really visible to me, so I'll make it more accurate, b, n plus 1. I hope this is well visible. So the second sum, first sum is multiplication of this by a. Second sum is multiplication of this same thing by b. Now if we will add them up together, we will basically have the total for a plus b to the nth and n plus 1, power of n plus 1. Now why did I position it this way? So it's easier to add because if I will start adding, I will add obviously elements with the same powers a and b. And all I have to do is I have to add these coefficients. All right, now I will wipe out this, I don't really need it anymore. And let me try to express this in a more concise way, equals. I still have n 0 a to the n plus 1 times b to the power of 0. Now, every other member except the last one will have a sum of two coefficients. Now let's see what kind of coefficients we have. This is n over 1, n over 0. Next will be n over 2 and n over 1. So every one will be a pair of n to the 1 plus n to 0. The member will be n to the n, a to the n, b to the first. Let me just put the next one. It will be n over 2 plus n over 1. So a is decreasing in power and b is decreasing and b is increasing in power. Okay? Well, the last member will be n 0. Oh, actually, I think I made a mistake here. It's supposed to be n over n. The last member. So the one before last will be n over n plus n over n minus 1, a to the first, b to the n. And the final member is this one. I'll just transfer it here. Okay, this is our expression. Now, let me ask you this question. And this is supposed to be a plus b to the power of n plus 1. Now, if you will go back to the a plus b to the power of n, you will see that it's basically very close to this one, except the formula which we would like to prove would be n plus 1, 0, a to the n plus 1, b to 0. Next would be n plus 1, 1. Now, this is a real formula which represents binomial theorem by Newton. This is a representation of the same original formula, the binomial theorem for n plus 1. These are coefficients which are increasing in the bottom part. And the sum of powers is always n plus 1. So this is the formula which we have to get, basically. It will start from a to the n plus 1 and b to the 0 and finish with a to the 0 and b to the n plus 1. But this is the formula which we got from using our mathematical induction supposition. So is it the same thing? Well, close, but not exactly. See, if I will compare, the members are exactly the same after the last one. This is the same with this one. The powers of a and b are the same. The coefficients are different, obviously. But now, here is the most important question. Are these coefficients really one and the same? Well, they are. And that's exactly what remains to be proven as the last step in this particular theorem. So let's think about the beginning and the end first. n over 0 and n plus 1 over 0. What are these coefficients? Now, by the way, do you remember that in case of a plus b square, the first coefficient with a square was 1? And the last one was also 1 with b square. Very similarly, with a plus b to the third degree, we also had the first member, a cubed, with the coefficient of 1. And the last one, b cubed was with the coefficient of 1. So it's not a coincidence. Actually, what is important is that any number, let's use different letters right now. Let's say m over 0. Let's use this. What is m over 0? It's m factorial divided by 0 factorial and m minus 0 factorial. Well, m minus 0 is m, obviously. So m factorial and m factorial is reduced. We will have 1. So for any m, m over 0 is 1. So that's why this coefficient is equal to this one because they are different in the top part, top number, and the bottom is 0. When it's 0, the whole coefficient is always 1. Now, very similarly, let's compare the last members, a to the power of 0 and b to the power n plus 1. Coefficient is n over n and pure coefficient is n over n plus 1. Now, what happens in this particular case? Well, let's just prove another theory that m over m is equal to m factorial divided by m factorial, m minus m factorial. Now, this is 0 and this is 0 factorial. I mean, so this is 1 and m over m factorial is the gain 1. By the way, I didn't mention here 0 factorial by definition is 1. You see, we were defining m factorial as 1 times 2 times 3, et cetera, up to n. What about 0? 0 is not defined. This is the definition of 0. Just forgot to tell. So that's why I'm just reducing it because it's 1. So as you see, the first member, n over 0 or n plus 1 over 0 is always 1. So they are equal. Now, the last members, n over n and n plus 1 over n plus 1, these are always equal. This is 1 and this is 1. So the last members are always equal. So what remains to be proven is that a sum of these two is equal to this, sum of these two is equal to this, et cetera. I mean, this one is equal to this. So let's do that. So what we have to prove is that n over 1 plus n over 0 is equal to n plus 1 over 1. Similarly, next one, n over 2 plus n over 1 should be equal to n plus 1 over 2. Similarly, up to the last one. Last one in this case is this and this, et cetera. So n over n plus n over n minus 1 should be equal to n plus 1 over n. So we have to prove all these formulas. If we will prove these, that will conclude the proof that this is equal to this and that actually means that we have proven completely that using our mathematical induction that a plus b to the power of n is represented by a normal formula of Newton. And let me just leave this very last thing on the board because this is the last but very easy, by the way, step. Now, instead of proving every one of these guys, I will just prove one which is a general case in this particular case. I will prove that m over k plus m over k minus 1. So k would be in this case 1 and m is equal to m is equal to m plus 1 over k. In this case, m is n and k is 1. m is n, k minus 1 is 0. m plus 1 is n plus 1 and k is still 1. In this case, m is n and k is 2, k minus 1 is 1. m plus 1 is n plus 1 and k, k is equal to the same 2. The last one. m is n and k is also n. Then m is n and k minus 1 would be the same as n minus 1. m plus 1 is n plus 1 and k is the same as this coefficient. So I will prove this formula. This is the most general formula which represents any one of those. If I will prove it, then basically ends up the whole story. So I don't need this. And to prove this one, I will just expand with factorials which is very easy. So let me just do it this way. m factorial divided by k factorial and m minus k factorial plus m factorial over k minus 1 factorial times m minus k minus 1 would be m plus 1 minus k factorial. That's on the left. Now, what do we do? Well, we obviously make it a common denominator in this particular case. How do we do it? Well, look, this is a multiplication of all the numbers from 1 to k. This is a multiplication of all the numbers from 1 to k minus 1. So if I will multiply by k, it will be k factorial, right? k minus 1 factorial is equal, sorry, multiplied by k is k factorial. It's by definition of the factorial. This is from 1 to k minus 1 plus k would be from 1 to k, the product of the number from 1 to k. So I will multiply this number by k, by k here and by k here. And this one together will be k factorial. Now, what is this? This is m plus 1, m minus k plus 1, if you wish. It will be a little easier. Minus k plus 1. So this is m minus k. This is m minus k plus 1. So again, this is the next factorial after this. So if I will multiply m minus k times m minus k plus 1, I will have this factorial. So here I will multiply it by m, m minus k plus 1, denominator and denominator, m minus k plus 1. Now, why did I do it? Because multiplication of m minus k plus 1 and m minus k factorial will give m minus k plus 1 factorial. These are all the numbers from 1 to m minus k. This is the next one. We'll give you the next one's factorial. Now, it's a common denominator, as you see. So I can put this equal to... All right, let's start from the denominator. In the denominator, in both cases, I have k minus 1 factorial times k will be k factorial, which is the same as this one. And m minus k plus 1 times m minus k factorial will be m minus k plus 1 factorial. So I will have k factorial times m plus 1 minus k factorial. I have reversed again here for convenience, you'll see why. Now, great. What do I have here? In the numerator, I can write it this way, right? m factorial times m minus k plus 1 and minus k plus 1 and plus k. m factorial is the common multiplier, so I just put it outside of the brackets, m minus k plus 1 and k. Well, obviously, we reduce this. And what do we have? m plus 1. Now, if we multiply m factorial and m plus 1, we will get m plus 1 factorial, right? And here we have k factorial and m plus 1 minus k. What is this? That's what it is. It's by definition, m plus 1 factorial over m plus 1 factorial and m plus 1 minus k factorial, right? This is the definition of the factorial. Instead of letter m, I'm using m plus 1. That's exactly what I have here. That's the definition of the final node in the whole proof. Because if you remember, we got two different sums, which had exactly the same elements. But the coefficients in one were these and in another, these. And since they're equal to each other, we basically have proven that the whole formula is expendable from m to m plus 1. And since we have already checked it for n equals to 2 and n equals to 3, well, we did it in explicit format. So let me just make sure that explicit format, which we had, really corresponds to the binomial formula of Newton. And that would be, remember, let's start with a plus b squared. What did we say? From the straight calculations, we had this. But let's use the formula, binomial formula of Newton, which is a plus b to the n's is n 0 a n b 0 plus n 1 a n minus 1 b 1 plus, etc. Plus 2 1 will be n n minus 1 a first b n minus 1 plus n over n a 0 b n. Okay, this is the general formula. Is it true for n equals to 2? Well, first of all, if this formula contains n plus 1 member, right, from 0 to n, which is n plus 1 member. Well, in this case, n is equal to 2, n plus 1 is 3. Yes, indeed, we have 3 members. Now, let's compare the coefficients. So 2 over 0 is equal to 2 factorial over 0 factorial 2 minus 0 factorial, obviously 1. Because 0 factorial is by definition 1, 2 minus 2 is 2, so we have exactly the same thing. Next one, 2 over 1 is 2 factorial over 1 factorial and 2 minus 1 factorial is equal to 1 factorial is 1, because it's multiplication product that holds the numbers from 1 to 1, so it's 1. 2 minus 1 is 1, so it's 1 again. And 2 factorial is 1 times 2, so the whole thing is equal to 2. 1, 1, 2. And the third one, obviously, will be the same thing. 2 factorial over 2 factorial times 2 minus 2 factorial. This is 0, 0 factorial is 1 by definition, and this is 1 as well, and this is 1. So, that corresponds. Now, it's not necessary to do anything else, because we have checked the formula for n equal to 2. Assuming that the formula is true for n, we switch to n plus 1, and we came up with exactly the same formula for n plus 1, so the proof is finished. But just to be on the safe side or to satisfy maybe certain curiosity, let's check as per 3, just in case. For a plus b to the third degree, if you remember in the very beginning of this lecture, we had a to the third plus 3 a squared b plus 3 a b squared plus b cubed. That's what we had. Well, let's check. Well, first of all, remember, power of n, n plus 1 elements in the sum. So, power of 3, n plus 1, which is 4, we have 4 elements added together. All right, coefficients obviously have to be checked, but the powers, we already see that if n is equal to 3, obviously it will be 3, 0, 2, 1, 2, 1. Next will be 1, 2, 1, 2, and the last power will be 3. So, all we have to do is to check the coefficients. The first one, n over 0 for whatever n doesn't really matter, it's always 1, so it's okay. For n over 1 with n is equal to 3, it will be 3 factorial over 1 factorial times 3 minus 1 factorial, which is what? Which is 3 factorial over 2 factorial, which is 1, 2, 3 over 1, 2, by definition of the factorial, which is 3, got this 3. Next one is n over 2, which is 3 factorial over 2 factorial and 3 minus 2 factorial equals 2. Same thing, 1, 2, 3, that's 3 factorial, 2 factorial is 1 times 2, 3 minus 2 is 1, so it's 1, this is 3, again. And the last one is 3 over 3 is always 1, again. So, that proves 2. So, as we see, it checks. Do you want to do it by 4? Well, I really don't want to do it, but again, out of curiosity, I'll do it by 4. 4 is equal to a plus b squared times a plus b squared, so let's just do it quickly here. a squared plus 2ab plus b squared times a squared plus, that's what happens when you want to do it fast, 2ab plus b squared equals 2. Well, let me be a little bit more scientific in this case. I know that whatever I'm multiplying, I will always have the combination of a to some degree and b to some degree, and since always every member has a sum of degrees equal to 2 in this case, and 2 in this case, when I multiply a member by any member, I will have a and b together in one element with powers, which if you add them together, you will always get 4, like a squared times a squared will be a to the fourth degree, or ab times b squared. This is first and the first, and this is the second degree, so the sum of degrees, sum of powers will always be 4. Well, if I know that the sum of powers will be always 4, I can say that this is sum coefficient with a to the fourth plus sum coefficient a to the third d, sum coefficient a to the second d to the second plus sum coefficient with ab cubed and sum coefficient with b to the fourth degree. And now the question is, what are the coefficients? Well, let me just guess it very simply. A to the fourth, I can get only if I multiply a squared by a squared, so this would be 1. Now, a cubed b, I can get either multiplying a squared by 2ab, so the coefficient is 2, or 2ab times a squared. I will also get a cubed b, so that's another 2. That's 4 together. Is there any way I can get another a cubed b? No, there is no way, so this would be 4. How can I get a squared b squared? Well, I can multiply a squared by b squared, or b squared by a squared, or by 2ab times 2ab. Now, a squared times a squared gives me, sorry, a squared times b squared gives me 1, a squared b squared. b squared times a squared gives me another, which is 2 together, and then 2ab times 2ab will be 4, a squared b squared, so it will be 6 all together, so that's my next coefficient, 6. How can I get a b squared? Well, I can't get it from here. I can get it either from here, which is 2ab times b squared, so I will get a b cubed. I'm looking for a b cubed, so a b cubed I can get from here and here. That's 2, or from here, this times this, which is another 2ab cubed, 4 all together. How can I get b to the fourth degree? Only by multiplying these two, so this is once again. Alright, so this is my formula. a plus b to the fourth degree is a to the fourth, 4ab cubed b, 6a squared b squared, 4ab cubed and 1b to the fourth degree. Well, let's check it. Number of elements in this sum is obviously m plus 1, which is 4 plus 1, which is 5, 1, 2, 3, 4, 5, and now let's check the coefficients. Now, 4 over 0, as usually is equal to 1, there is no problem with this. Now, 4 over 1, 4 factorial divided by 1 factorial and 4 minus 1 factorial, which is what? 1, 2, 3, 4 divided by 4 minus 1 is 3, 1, 2, 3. Obviously, this is reduced because 4. Get this one. Next, 4 over 2, 4 factorial over 2 factorial and 4 minus 2 factorial. 1, 2, 3, 4 divided by 2 factorial is 1 times 2 and 4 minus 2 is 2 again. 1 minus 1 times 2. So what do we have here? 2 and 2 is 4, so we can reduce by 4. We have 2 times 3. 6, 6. 4 over 3, 4 factorial over 3 factorial, 4 minus 3 factorial is equals 2. 1, 2, 3, 4 over 1, 2, 3. This is 1. Reduced, we have 4. We have 4. And the last but not least, 4 over 4 factorial is 1, which we have proven many times before. Well, that's one. As you see, the formula works and the first three elements, 4, 3, formulas for n is equal to 2, 3, and 4, we even derived in a real explicit way and they correspond to by a nominal formula. Well, that's it. That's the end of the proof. Thank you very much.