Mathematical Induction for an alternating series It's been a while since I have done a problem like this. I have this problem

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

So I have gotten this far:

Base Case: 
$$n=1$$

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

Assume:

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

Prove:
$n = n+1$

$$\sum_{i=0}^n+1 (-1)^ {i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I think that I can just replace the i's with (n+1) too

Which would be this

$$\sum_{i=0}^{n+1} (-1)^ {n+2} (n+1)^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I don't remember where to go from here.
You can most certainly not replace $i$s with $n$s. That's a major mistake. For example

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
while 
$$\sum_{i=1}^n n = n^2$$ and those are certainly not the same thing.



What you do have to do is replace $n$ with $n+1$ everywhere, including the sum, so what you have to prove is 

$$\sum_{i=0}^{n+1} (-1)^{i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2}$$

while you can assume that 

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

is true.

To prove the above equation, remember that 

$$\sum_{i=0}^{n+1} a_i = \left[\sum_{i=0}^{n}a_i\right] + a_{n+1}$$

and now use this fact on your particular sum, then use the induction assumption that tells you something about the sum in parentheses.
See the summation of the n+1 terms is just the summation of n terms (which is known) + $(-1)^{n+2}(n+1)$. Now you see the minus sign $(-1)^{n+1}$can be taken common the remaining term is :

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

Thus at last the minus sign turns out to e $(-1)^{n+2}$ and hence proved.