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# Proof of finite arithmetic series formula

Watch Sal prove the expression for the sum of all positive integers up to and including n. Created by Sal Khan.

Video transcript

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.