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??

Notice $$n^2 = \frac{(n-1)n + n(n+1)}{2}
\quad\implies\quad (-1)^{n-1} n^2 = (-1)^{n-1}\frac{(n-1)n}{2} - (-1)^{n}\frac{n(n+1)}{2}$$
The sum is a telescoping sum and
$$\require{cancel}\sum_{k=1}^n (-1)^{k-1} k^2 = \color{red}{\cancelto{0}{\color{grey}{(-1)^{1-1}\frac{(1-1)1}{2}}}}- (-1)^n \frac{n(n+1)}{2}
= (-1)^{n-1}\frac{n(n+1)}{2}
$$
