# Thread: Proving a sum is equal to an expression for all positive integers n

1. ## Proving a sum is equal to an expression for all positive integers n

Prove that ∑ from k =1 to n of (-1)^k*k^2 = (-1)^n*((n(n+1)/2) for all positive integers n.

In words: Prove that the sum (from k =1 to n) of negative one to the k times k squared is equal to negative one to the n, times (n times (n plus 1), divided by 2.

I am beyond lost.

2. ## Re: Proving a sum is equal to an expression for all positive integers n

beyond lost there is proof by induction

3. ## Re: Proving a sum is equal to an expression for all positive integers n

Do you know what "proof by induction", mentioned by Idea, is?

Proof by Induction follows from a basic property of the positive integers- starting from 1, we can get all positive integers by adding 1 repeatedly

To prove Statement P(n), which depends upon the positive integer, n (In your problem, P(n) is the statement $\sum_{k=1}^n (-1)^k k^2= (-1)^n \frac{n(n+1)}{2}$.

So first prove P(x), which, since the sum from 1 to 1 has just one term, is $(-1)^1 1^2= (-1)^1\frad{1(1+1)}{2}$. That's obviously true- the left side is -1 and the right side is $(-1)\frac{1+ 1}{2}= -1$.

Now prove that "If P(m) is true then P(m+1) is true". (I prefer not to use "n" here- that makes it look like we are assuming what we want to prove. "n" refers to all positive integers. Here, "m" is a specific integer.)

That is assuming $\sum_{k=1}^m (-1)^k k^2= (-1)^m \frac{m(m+1)}{2}$ we want to prove that $\sum_{k=1}^{m+1} (-1)^k k^2= (-1)^{m+1} \frac{(m+1)(m+2)}{2}$.

The sum on the left of that last statement is just the sum on the previous statement [b]plus[b] one additional term:
[tex]\sum_{k=1}^{m+1} (-1)^k k^2= $\sum_{k=1}^{m} (-1)^k k^2+ (-1)^{m+1}(m+1)^2$. By the "assumption" we can write that as [tex]\sum_{k=1}^{m+1} (-1)^k k^2= $(-1)^m\frac{m(m+1)}{2}+ (-1)^{m+1}(m+1)^2$

Now add the terms on the right side and show that you get $(-1)^{m+1}\frac{(m+1)(m+2)}{2}$.