How to prove this by induction

Question

Prove by induction the following equality :

$\ 1-4+9-16+\cdots+(-1)^{n+1} n^2 = (-1)^{n+1}(1+2+3+\cdots+n) $

I don't know what to do in this case, I know what to do in general but can do this one

Answer 1

For inductive step :

Suppose that $$1-4+9-16+\cdots+(-1)^{n+1}n^2=(-1)^{n+1}(1+2+\cdots+n).$$ Then, $$1-4+\cdots+(-1)^{n+1}n^2+(-1)^{n+2}(n+1)^2$$$$=(-1)^{n+1}(1+2+\cdots+n)+(-1)^{n+2}(n+1)^2$$ $$=(-1)^{n+1}\cdot\frac{n(n+1)}{2}+(-1)^{n+2}(n+1)^2$$ $$=(-1)^{n+2}(n+1)\left(-\frac n2+n+1\right)$$$$=(-1)^{n+2}\frac{(n+1)(n+2)}{2}$$$$=(-1)^{n+2}(1+2+\cdots+n+(n+1)).$$