Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?

No one's pointing out my favorite result that $n^2$ is the sum of the first odd numbers?
$(n + 1)^2  - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1$.
and the result follows...
Case 1: $k=2m$ is even.
$1 - 2^2 + 3^2 - 4^2 + .... + (k-1)^2 - (k)^2 = -(2^2 -1) - (4^2 - 3^2) - ...- (k^2 - (k-1)^2)=$
$-(3) - (-7)-....-(2k + 1) = \sum_{j = 1}^m-(4j + 1)=$
$-4(\sum_{j=1}^m j) + m(\sum_{j=1}^m j)=$
$-4(\frac{(m(m+1)}{2}) + (m) = -2m(m+1) + m = m(-2(m+1) + 1) = m(-2m - 1) =$
$ -m(2m + 1)= -\frac{k(k + 1)}{2}$
Case 2:  $k = 2m + 1$ is odd.
$[1 - 2^2 + 3^2 - 4^2 + .... + (k-2^2 - (k-1)^2 ] + k^2 =$
$ -\frac{(k-1)k}{2} + (k^2) = k[-\frac{k-1}{2} + k] =$
$k[\frac{-(k-1)+2k}{2} ]=\frac{k(k + 1)}{2}$
So 
sum = $(-1)^{k-1}\frac{k(k + 1)}{2}$
