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?

It can be easily shown that the series on the left reduces to the sum of integers, multiplied by $-1$ if $k$ is even.
For even $k$: 
$$\begin{align}
&1^2-2^2+3^3-4^2+\cdots+(k-1)^2-k^2\\
&=(1-2)(1+2)+(3-4)(3+4)+\cdots +(\overline{k-1}-k)(\overline{k-1}+k)\\
&=-(1+2+3+4+\cdots+\overline{k-1}+k)\\
&=-\frac {k(k+1)}2\end{align}$$
Similar, for odd $k$, 
$$\begin{align}
&1^2-2^2+3^3-4^2+5^2+\cdots-(k-1)^2+k^2\\
&=1+(-2+3)(2+3)+(-4+5)(4+5)+\cdots +(-\overline{k-1}+k)(\overline{k-1}+k)\\
&=1+2+3+4+\cdots+\overline{k-1}+k\\
&=\frac {k(k+1)}2\end{align}$$
Hence, 
$$1^2-2^2+3^3-4^2+5^2+\cdots+(-1)^{k-1}k^2=(-1)^{k-1}\frac {k(k+1)}2\quad\blacksquare$$
