Induction of $\sum_{k=1}^{n} (-1)^{n-k}k^2 = \frac{n(n+1)}{2}$

Question

I'm trying to prove following statement through induction:

$\sum_{k=1}^{n} (-1)^{n-k}k^2 = \frac{n(n+1)}{2}$

I have only seen how to prove with induction when the variable $n$ is not included in the sum function. Like here (this statement is not correct, just to provide an example):

$\sum_{k=1}^{n} (-1)^{k}k^2 = \frac{n(n+1)}{2}$

It confuses me and I don't know how to proceed. Any advice?

Thank you.

Answer 1

$$\sum_{k=1}^n(-1)^{n-k}k^2=(-1)^n\sum_{k=1}^n(-1)^{-k}k^2=(-1)^n\sum_{k=1}^n(-1)^kk^2\\\sum_{k=1}^n(-1)^kk^2=(-1)^n\frac{n(n+1)}{2}$$

Answer 2

We can derive from scratch as well. For even $n=2 m$, we can simply pair up the summands to get

$$\sum_{k=1}^{m} ((2 k)^2 - (2 k-1)^2) = \sum_{k=1}^{m} (4 k-1) = 2 m (m+1) - m = \frac{n (n+1)}{2} $$

For odd $n=2 m+1$, we may do the same, but we have a last (positive) term, and we change the sign of the even part of the sum. Thus we get

$$(2 m+1)^2 - 2 m (m + 1) + m = 2 m^2 + 3 m + 1 = 2 \left (\frac{n-1}{2} \right )^2 + 3 \left (\frac{n-1}{2} \right ) + 1 = \frac{n (n+1)}{2}$$