Sum of terms $1^2-2^2+3^2-...$ Prove that: $$1^2-2^2+3^2-4^2+...+(-1)^{n-1}n^2=\frac{(-1)^{n-1}n(n+1)}{2}$$
I proved it by induction but is there any other way to solve it?
If it was not a proof but rather a question like find the term,how to solve it?
I realized that alternate terms were under same sign but can't understand whether to take $\frac{n}{2}$ odd and even terms[if n is even] or $\frac{n+1}{2}$ odd terms and $\frac{n-1}{2}$ even terms[if n is odd]. 
I thought of this $$1^2-2^2+3^2-4^2+5^2=1^2+{(1+2)}^2+{(1+4)}^2-2^2-4^2$$
$$=1^2+1^2+1^2+2(0+2+4)+2^2+4^2-2^2-4^2$$
But then how to generalize??

Here's a nice trick. For each $k$,
$$k^2=\frac{(k-1)k}2+\frac{k(k+1)}2=a_{k-1}+a_k$$
say.
Then
$$1^2-2^2+3^2-4^2+\cdots\pm n^2
=a_0+a_1-a_1-a_2+a_2+a_3-\cdots\pm a_{n-1}\mp a_n$$
and almost all of these terms cancel.
