1. ## Summation Induction help

Hello, I've read everything I can on induction for the most part it makes sense, but I'm having trouble with this summation problem:

SUM from i = 1 to n: (-1)^(i+1)(i^2) = [ (-1)^(n+1) n(n+1) ] / 2

(I dont know latex...yet)

Basis step:
1 = 1 check

Inductive step:
plugging in n+1 and trying to solve I can't seem to get anywhere...
Please point me in the right direction

Thank-you for any help!!
jeremywho

2. Well, I'll assume you know how induction is done and how it works. It's just a matter (in this case) of calculation. In fact, suppose the equality holds for some $n \in \mathbb N$, that is:

$\sum_{i=1}^n (-1)^{i+1}i^2=(-1)^{n+1}\frac{n(n+1)}{2}$

You wish to prove it also holds for $n+1$, hence you want to prove that:

$\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}\frac{(n+1)(n+2)}{2}$

Simply observe that:

$\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (-1)^{i+1}i^2$

and recall that you are assuming the equality is true for $n$ (which means you know the corresponding expression for that sum on the right side).

Hope it helped

3. Thank-you so much for the reply!!

I understand we want to prove this:
$
\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}\frac{(n+1)(n+2)}{2}
$

I don't understand how we get:

$
(-1)^{n+2}(n+1)^2
$

in

$
\sum_{i=1}^{n+1} (-1)^{i+1}i^2=(-1)^{n+2}(n+1)^2+\sum_{i=1}^{n} (-1)^{i+1}i^2
$

Maybe its just simple algebra I'm missing?

Thank-you Again!!

4. You just pick the last term of

$\sum_{i=1}^{n+1} (-1)^{i+1}i^2$

which is $(-1)^{n+2}(n+1)^2$ and write it independently. A straightforward, numerical example:

$\sum_{i=1}^{5}i = 1+2+3+4+5 = 5+(4+3+2+1) = 5+ (\sum_{i=1}^4i)$

The sums don't have the same upper bound! One ends in $n+1$ and the other in $n$.

5. ## Got it!

I finally got it! I just wanted to thank-you again for your help!