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??

For example, if $n$ is odd, $n=2k-1$ then we obtain:
$$LS=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{k(k-1)(2k-1)}{6}=$$
$$=\frac{n(n+1)(2n+1)}{6}-8\cdot\frac{\frac{n+1}{2}\cdot\frac{n-1}{2}\cdot n}{6}=\frac{n(n+1)}{2}$$
