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?

Proceed inductively. Verify that for $k=1$, $1 = 1$. 
Now suppose the result holds for $n - 1$. Then 
$$ 1^2 + \dots + (-1)^{n-1} n^2 = 1 + \dots + (-1)^{n-2} (n-1)^2 + (-1)^{n-1} n^2 = (-1)^n \frac{n(n-1)}{2} + (-1)^{n-1} n^2 = (-1)^n \left(\frac{n(n-1)-2n^2}{2}  \right) = (-1)^n \left(\frac{-n^2-n}{2} \right) = (-1)^{n-1} \left(\frac{n(n+1)}{2}\right)$$
The induction hypothesis was applied at the second equality. 
