Help with proof by induction problem Any input on this problem I'm working on?
I want to prove by induction that for all natural $n$ we have that 
$\sum_{i=1}^n -1^jj^2 = (-1)^n\frac{n(n+1)}{2}$
So far I have:
1) Proved the base case, n = 1, which gives me: 
$\sum_{i=1}^n -1^jj^2 = (-1)\frac{2}{2}$
$\rightarrow\sum_{i=1}^n -1^jj^2 = -1$
Which makes sense.
2) For the induction step, I assume n = k is true, I get:
$\sum_{i=1}^n -1^jj^2 = (-1)^k\frac{k(k+1)}{2}$
Then I want to find out if this is true for n = k+1. I insert for k+1.
$\sum_{i=1}^n -1^jj^2 = (-1)^k+1\frac{k+1((k+1)+1)}{2}$
I know you guys usually ask for me making an effort at it, but I am literally stuck, and have been so for some time. Any tips?
BTW I have read this guide, which was a good read, but didn't really help me figuring out how to solve these. Please share other links if you have any.
 A: Hint:
$$\sum_{j=1}^{k+1}(-1)^j j^2 = \left[\sum_{j=1}^{k}(-1)^j j^2\right] + (-1)^{k+1}(k+1)^2.$$
Now you can use the assumption on the bracketed term.
You may need to consider even and odd $k$ separately.
(Note:  I'm not sure $-1^k$ and $(-1)^k$ are the same thing.  The second one makes it clear what you mean.)
A: Here you are,
Assume:
$\sum_{i=1}^k -1^jj^2 = (-1)^k\frac{k(k+1)}{2}$
Using the hypothesis show that:
$\sum_{i=1}^{k+1} -1^jj^2 = (-1)^{k+1}\frac{k+1((k+1)+1)}{2}$
the part you are missing:
$\sum_{i=1}^{k+1} -1^jj^2 = (\sum_{i=1}^{k} -1^jj^2) + (-1)^{k+1}(k+1)^2$
Now substitute based on the inductive hypothesis.  Apply your algebra skills and take it home.
A: $$\sum_{j=1}^{k+1}(-1)^j j^2=(-1)^{k+1}(k+1)^2+\sum_{j=1}^{k}(-1)^j j^2=(-1)^{k+1}(k+1)^2+(-1)^k\frac{k(k+1)}{2}$$
therefore
$$\sum_{j=1}^{k+1}(-1)^j j^2=(-1)^{k+1}\left((k+1)^2-\frac{k(k+1)}{2}\right)$$
$$\qquad=(-1)^{k+1}\frac{(k+1)(k+2)}{2}$$
A: When dealing with induction proofs, the usual strategy is to try to find some way to reduce the expression we get when inserting $n=k+1$ to the expression when $n=k$ so we can apply the assertion given by the inductive step and then work from there.
If the expression in question is a sum (as it is the case) it is straightforward to to just that, as we can pop the last term of the summation to get the expression when $n=k$.
Thus:
$$
\sum_{i=1}^{k+1} (-1)^jj^2 = (-1)^{k+1}(k+1)^2 + \sum_{i=1}^{k} -1^jj^2= (-1)^{k+1}(k+1)^2 + (-1)^k\frac{k(k+1)}{2}
$$
Now it is a matter of applying algebra.
