Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College. Mathematical induction. Let SN be a statement involving the positive integer N If first S one is true and 2nd the truth of the statement S sub K implies the truth of the statement S sub K + 1 for every positive integer K, then the statement SMN is true for all positive integers N. Generic even number is of the form 2M where M is an integer. So 2 * 1, two times 2 two times three 2 * 4 are all examples of even numbers. An odd number is one more than an even SO2M plus one, where M is an integer. We're always going to simplify S sub K&S sub K + 1 before doing induction. Now when we're looking at this, we have our A sub one term, our A sub two term. Somewhere out here we're going to have an A sub K -, 1 term, an A sub K term, an A sub K plus one term all the way out to our A sub N term. So when we talk about S sub K, we're talking about the sum of the first K terms. When we're talking about S sub K + 1, we're talking about the sum of the 1st K + 1 terms. So if we can prove it's true for the first one, and then if we assume it's true for the first K terms prove that it's true for the K + 1. If we had know that it's true for the first one and we let K be 1, then we can prove that it's true for the second. And then if it was true for the second, we can prove for the third. If it's true for the third, we could prove for the 4th. That's why we choose S sub K&S sub K + 1. They're somewhere within the series. Let's look at a few examples. If we have 2 + 4 + 6 plus... Plus 2N equaling n * n + 2. So the first thing we're going to do is show that it's true for S sub one. Does 2 the first term equal 1 * 1 + 1, so 2 does equal 2. If we wanted to show a few more terms we could say does S sub two or the sum of the 1st 2 terms equal on the right side we would put 2IN for our N. So does 6 equal 2 * 3? The answer would be yes. If we wanted to show S sub three, we would add the first 3 terms and we would say does that equal 3 * 3 + 1 and 2 + 4 + 6 is 12 and 3 * 4 is 12. So the first 3 terms definitely check out. So what we're going to do next is we're going to let it be true for the first K terms and then I personally leave some room. And what we want to do is we want to prove that it's true for the K plus one term. So if we look and simplify things up first on this K + 1, we'd get K + 1 * K + 2. If we look at our left hand side, this left hand side is almost exactly the same as the K + 1 left side. The only thing that's different is this last term here. So we're going to add that term to each side. We're going to add it to the left, and we're going to add it to the right. So 2 + 4 + 6 + + 2 K plus 2 * K + 1. Now my left side is exactly the same. If I have my left side, though, I have to do the same thing to my right side. So I'm going to add that 2K plus one to each side. Now on the right side, if I thought about distributing, combining my like terms and then factoring, I can see that this does indeed give me what I needed it to give me on the right side. So we started with the K term being true. We made an assumption this is true. We added the K plus one single term to the left, right side and right side, and we showed that it was true for the S sub K + 1. Let's look at another one, 1 ^3 + 2 ^3 plus... plus N cubed. So we're going to show it's true for S sub one. The first term on the left has to equal on the right when we stick in our one. So does 1 equal 1 * 2 ^2 / 4? The answer is yes, we don't have to. But if we wanted to show a couple more terms just to get the idea, we'd have 1 ^3 + 2 ^3 equaling 2 ^2 2 + 1 ^2 / 4. So does 1 + 8 equal 4 times 9 / 4? And the answer is yes, 9 does equal 9. Now if anywhere in our S1 or S2 or S sub three, we get a no. We don't have to go any further because we know it won't hold. It won't be true for mathematical induction. So we start the next part with making the statement that the S sub K term is true. It's important that we have that statement made. If it's true for this term, then we're trying to prove that it's going to be true for the S sub K plus one term. And literally we just stick in K + 1 for that N in the original. So if I simplify this up, I'd get K + 1 quantity squared, K + 2 quantity squared all over 4. Now our thought process should be the left sides are exactly the same. Actually, the left sides are not exactly the same yet we're going to add this K + 1 ^3 term to each side. Then the left sides would be exactly the same, so 1 ^3 + 2 ^3 + + K ^3 + K + 1 ^3 = K ^2, K + 1 ^2 / 4 plus the K + 1 cube. So now my left side is exactly what I needed it to be. I have to manipulate the right side to get it to be AK plus 1 ^2 * K + 2 ^2 / 4. The first thing we're going to do is we're going to get a common denominator. So the second term is going to turn into 4K plus one cubes over 4. Now we can see that both terms have a four in common. They also both have AK plus one squared. So if I thought about taking K + 1 ^2 out of the numerator, if I took AK plus one squared out of this first term, that would leave me just AK squared. If I take AK plus one squared out of the second term, that would leave me a 4K plus one. So now what we're going to do is we're going to distribute K ^2 + 4, K plus 4 all over four. Well, K ^2 + 4 K plus four factors, and it factors into K + 2 ^2. And that's what we were trying to get it to equal. So we've just proven by mathematical induction that this is true. Let's look at the next. 1-2 is a factor of n ^2 + n So for S sub one, is 2 a factor of 1 ^2 + 1? And the answer is yes. So now we want to show if 2 is a factor of K ^2 + K, then prove that 2 is a factor of K + 1 ^2 + K + 1. So if we look at this K + 1 ^2 + K + 1, that's really K ^2 + 2 K plus 1 + K + 1, or K ^2 + 3 K +2. So I need to figure out how to show that it's really a factor of K ^2 + 3 K +2. So we start with two is a factor of K ^2 + K. We start with that S sub K. Well what do I need to do to add to K ^2 + K? If my goal is to get that K ^2 + 3 K plus two, I have to add a 2K and A2. So what I really need to do is I need to show that that 2K plus 2 is a factor of 2. We knew that this was a factor of 2 that was given. If I factored out a two of the K + 1. Is that second part a factor of 2? Yep, because two times anything is a factor of 2. Two times 1/2 times two 2 * 3. And the fact that the K ^2 + 3 K plus 2 is really just K + 1 ^2 + K + 1 shows that this is indeed a true statement. 2 is a factor of anything of the form n ^2 + n Use mathematical induction to prove the statement for all positive integers N So we're going to show it's true for S sub 1/4 equal 1 * 5 * 1 - 1, so 4 equal 4 true. We're going to make the belief, the statement that it's true for the KTH term, and then we're going to prove that it's true for the K + 1 TH term. If we simplify this up, we get K + 1 * 5 K plus 4. So our very first step is going to take and add the K plus one term to each side. So now the left side is exactly what I needed it to be, and we're going to manipulate the right side to show that that's really K + 1 * 5 K plus 4. So if I distribute, I get five K ^2 -, K + 10 K plus 10 -, 6 five K ^2 + 9 K plus 4, and that's going to factor into K + 1 five K + 4, which is what we were trying to get. And hence by mathematical induction, we've just proven this. Thank you and have a wonderful day.