Find the sum $\sum\limits_{k=1}^{2n} (-1)^{k} \cdot k^{2}$ How to find this sum?
$$\sum\limits_{k=1}^{2n} (-1)^{k} \cdot k^{2}$$
 A: As I'm a fan of proofs without words, here is my pictorial effort of why
$$\sum_{k=1}^{2n} (-1)^k k^2 = \sum_{i=1}^{2n} i = n(2n+1).$$

A: Try working with the sequence $S_n=\sum_1^{2n}(-1)^kk^2$. If you look at the sequence of differences (sidrat hefreshim in hebrew, i'm not sure if that the correct translation. any hebrew speaker- please correct me here) of $A_n=S_{n+1}-S_n$ you should be able to get a first degree formula for $A_n$.
Now- Notice that $S_n-S_1=\sum_1^{n-1}A_n$, and since $S_1$ is easy to calculate, and $\sum A_n$ is not to tricky- this should give you the solution.
good luck!
A: Sorry, I am in a bit of a hurry, but it should be this sequence.
To see this, just remark that you are summing up difference of consecutive squares (-1+4) + (-9+16) + ..., and these difference are the odd integers congruent to 3 mod 4  (i.e., 3, 7, 11...).
A: Hint: What is $(2i)^2-(2i-1)^2$? Next, what are $\displaystyle\sum_{i=1}^n1$ and $\displaystyle\sum_{i=1}^ni$?
A: You can separate this sum into two, one for $k$ odd and another for $k$ even, writing $k=2t+1$ and $k=2t$, respectively. You can also try finding $\sum k^2 z^k$ and then set $k=-1$.
A: Yet another way (although not as simple as those already suggested) is to write it as
$$-\sum_{k=1}^{2n} k^2 + 2 \sum_{m=1}^{n} (2m)^2$$
and use the known formula for a sum of squares.
