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In the last video, we proved that the sum of all of the positive integers up to and including n can be expressed as n times n plus 1 over 2. And we proved that by induction. What I want to do in this video is show you that there's actually a simpler proof for that. But it's not by induction, so it wouldn't be included in that video. But I'll show you that it exists, just so you know that induction isn't the only way to prove it. So we define that function S of n as the sum of all of the positive integers up to and including n. So this is equal to, by definition, 1 plus 2 plus 3 plus, all the way to plus n minus 1 plus n. So it's the sum of all of the integers up to and including n. This is how we're defining it. Well, we can rewrite it again. We can say that the sum, S of n-- we could just rewrite this same thing, but we could rewrite it in a different order. We could say that this is the same thing as n plus n minus 1 plus n minus 2 plus, all the way down to plus 2 plus 1. Now, what does this do for us? Well, we can actually add these two rows. If we add S of n plus S of n, we're going to get 2 times this sum, so we're just adding on the left. And then we can also add on the right. So we're just adding this sum twice, but what's interesting is how we're going to add it. We're going to add this term to this term, this term to this term, because we're really just trying to add these two things. And we can pick any way we want to add them. So 1 plus n is going to be n plus 1. And then we're going to add-- let me do it in pink-- 2 plus n minus 1. So what's 2 plus n minus 1? Let me write it over here. 2 plus n minus 1. It's the same thing as 2 plus n minus 1, which is the same thing as n plus 1. 2 minus 1 is just 1. So this is also going to be n plus 1. And then this term over here, 3 plus n minus 2, or n minus 2 plus 3. Once again, that's going to be n plus 1. And you're going to do that for every term all the way until you get over here, n minus 1 plus 2. That's also going to be n plus 1. And then finally, you have n plus 1 right over here. Plus n plus 1. So what's this whole sum going to be? Well, how many of these n plus 1's do we have? Well, we have n of them, for every term in each of these sums. So this is 1, 2, 3, count all the way to n. You have n of these terms. So you have n n plus 1's. So if you add something to itself n times, or if you have something n times right over here, this is exactly equivalent to n times n plus 1. So 2 times that sum of all the positive integers up to and including n is going to be equal to n times n plus 1. So if you divide both sides by 2, we get an expression for the sum. So the sum of all the positive integers up to and including n is going to be equal to n times n plus 1 over 2. So here was a proof where we didn't have to use induction. It's really kind of a pure algebraic proof.