 I am Zor. Welcome to Unizor education. This is a continuation of the topic on mathematical induction. We'll talk about some problems. So this is the first problem. Here is the first problem. And we will use the method of mathematical induction to prove some very interesting formula. Look at the following. If you have number 1 and you square number 1, you will get 1. If you will get a sum of 1 and 3, the number of these numbers in this sum is 2. And if you will square 2, it will be 4. And if you know just, this is 4. Let's take the next odd number, 1, 3 and 5. It's 9. Number of these numbers in this sum is 3. And 3 square is 9 again. Well, is this a coincidence? Definitely not. And here is what we are trying to prove. That if we will have certain number of odd numbers added together, then the sum will be equal to the number of these numbers squared. And here is how to express it mathematically. If you have 1 plus 3 plus 5 plus 7, etc. This is member number 1. This is member number 2. This is member number 3. This is member number 4. So this is a number of the number of our sum. How can we connect the number of this number and its value? 1 and 1, 2 and 3, 3 and 5, 4 and 7. So how can we say that the number 4 in this sequence of odd numbers is 7? That's actually quite simply. It's 2 times n minus 1. This is a general formula for n is equal to 1. The first number is 1. If n is equal to 2, 2 times 2, 4 minus 1 is 3. If n is equal to 3, 3 times 2, 6 minus 1 is 5. And if n is equal to 4, 4 times 2 is 8 minus 1, 7. So this is the generalized formula which gives you the nth odd number. So what I'm trying to prove is that if I will have first n odd numbers and add them together, so the nth will be 2 times n minus 1. This is number n. So if I will have n odd numbers and sum them together, I will try to prove that this is n squared. Well, this is not really obvious. I mean, we did check it in a couple of initial cases. And how can we prove this formula in the general case? Well, obviously the method of mathematical induction is very handy in this case. That's why I'm presenting this as a problem. I think this is a very good position right now to basically pause the video and try to solve this problem just by yourself. So just press the pause button. Think about this particular problem. And then play it again so whatever I'm trying to prove right now is basically an explanation and solution to this particular problem. It would be interesting if you first try it and then check it against whatever I'm trying to prove. Alright, so now we will have a solution to this problem after you pause and try it yourself, obviously. So this is the problem. This is the formula which we are trying to prove. Okay, if you remember the method of mathematical induction consists of three steps. Step number one, we check for n is equal to 1. Well, let's check it. For n is equal to 1, I am supposed to have only one member on the left. n times 2 is 2 minus 1 is 1. So this is the generalized expression for the first member only. So n is equal to 1. On the left, I have 1. On the right, I have 1 square which is 1. Check. Fine, no problem. Now, let's assume that 1 plus 2 plus etc plus 2 times certain number k minus 1 is equal to k square. Let's assume this is true for some particular k. And we have to prove using this, of course, that 1 plus 2 plus etc plus the 1 before last member will do this. And if instead of n, I will put k plus 1, let's think what will be. k plus 1 times 2 minus 1. This is my formula. This is my sum on the left with k plus 1 members. Okay, let's transform it. Since we know that this piece by assumption is equal to k square, we can say that the whole sum is equal to k square plus, and let me just make it a little simpler. 2k plus 2 minus 1 is equal to 2k plus 1. Right? 2k plus 2 minus 1. k plus 1. Now, great. This is, by the way, k plus 1 square. If you don't believe me, just check it out yourself. This is the formula for k plus 1 square. No, indeed, for those who... Okay, I will check it for you. k square is k plus 1 times k plus 1, which is k times k k square. k times 1 is k. 1 times k is k. And 1 times 1 is 1. And this is 2k. 2k. k square plus 2k plus 1. All right. So what it looks like, that using an assumption for n is equal to k, I have proved that basically this formula, you see this k plus 1 square? That's exactly what it is. This is true for n is equal to k plus 1. So all three steps of mathematical induction are performed, validated, everything is fine. So the whole thing, therefore, is proven. So we can prove that for any n, this formula is true. Let me again repeat these logical steps, which have to be just said about this proof. Since we have checked it for 1, and we also assuming that the formula is true for some number k, it will be true for number k plus 1. We can really make a conclusion that since we checked for 1, for k equals 1, then it will be true for 2. We make a transformation from k to k plus 1. So from 1, we can make a transformation to 2. From 2 to 3, from 3 to 4, et cetera. So any number of odd numbers, sequential odd numbers added together, if their number is n, then it will be n square root of sum. That's a complete proof. Thank you very much.