# Mathematical Induction - Problem 1

###### Transcript

We’re now going to use mathematical induction to prove the sum of n introduced is equal to n times n plus 1 over 2.

Our first step is to show it works for n equals 1. Basically we’re assuming the first integer, obviously the sum of the first integer is just going to be 1. But we want to make sure that our formula holds. So when we plug in one here we get 1 times 1 plus 1, 1 times 2 is 2, over 2 which is equal to 1, so we have 1 is equal to 1, that works.

We now assume that it works for some arbitrary value k. What that tells us is we end up with 1 plus 2 plus 3 plus so on and so forth, all the way up to k, is going to be equal to k times it’s going to be k plus 1 over 2. So then sing this fact we want to show that it works for k plus 1. So what we can do is show it works for k plus 1.

What we’re looking for is the sum of 1 plus 2 plus 3 plus k plus k plus 1. What we have is this piece we already know to be k times k plus 1 over 2, because all we did is we added k plus to this side we therefore we add k plus 1 to that side as well. All we have to do in this case is to get a common denominator. In this case this is going to be 2. Multiply this by 2 over 2 and we’re left with, let’s get a different color so we can distinguish it a little bit more. Foiling this out k² plus k over 2 plus 2k plus 2 over 2. Combining like terms we end up with k² plus 3k plus over 2, which hopefully you can see factors down into k plus 1 times k plus 2 over 2.

So that is on one side of our equation. If we were to just add k plus 1 which is the number we added to this side, we would end up with k plus 1 times k plus 2 over 2. We want to prove that this statement is the same as this statement over here. For this statement we can just see our formula.

The sum of the first n terms is simply n times n plus over 2. So the sum of the first k plus 1 terms is simply our n 1 to n so simply k plus 1. We added one to that over here so that turns into k plus 2 over 2.

So basically by assuming, by proving it for one, assuming it works for k, we were then able to manipulate this equation to get the equation to then work for k plus 1, thus proving that this statement is true by mathematical induction.