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??

You could use formal manipulations with a truncated power series:
$$\frac{1-x^{n+1}}{1-x} = 1 + x + x^2 + \cdots + x^n \\
\frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = 1 + 2x + \cdots + n x^{n-1} \\
x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) = x + 2x^2 + \cdots + n x^n \\
\frac{d}{dx} \left( x \frac{d}{dx} \left( \frac{1-x^{n+1}}{1-x} \right) \right) = 1 + 4x + \cdots + n^2 x^{n-1}
$$
Now evaluate the derivatives on the left hand side, and then substitute $x := -1$.
