Find the sum $\sum_{r=1}^n (r(r+1))^2 (-1)^r$ in terms of n.

Question

Find the sum $$\sum_{r=1}^n (r(r+1))^2 (-1)^r$$ in terms of n.

My approach

I tried to make telescoping terms by adding another a constant term written in terms of r, $((r+2)-(r-1))$ but the $(-1)^r$ wouldn't allow them to cancelled.

Answer 1

When $n$ is even $$\begin{align*} \sum_{r=1}^n(r(r+1))^2(-1)^r&=\sum_{r=1}^{n/2}((2r+1)2r)^2-((2r-1)2r)^2\\ &=32\sum_{r=1}^{n/2}r^3\\ &=32(\frac{n^2}{8}+\frac{n}4)^2\\ &=n^2(n+2)^2/2, \end{align*}$$ using the classic formula $\sum_{i=1}^mi^3=(m(m+1)/2)^2$. Similarly if $n$ is odd $$\begin{align*} \sum_{r=1}^n(r(r+1))^2(-1)^r&=\sum_{r=1}^{n-1}(r(r+1))^2(-1)^r-(n(n+1))^2\\ &=(n-1)^2(2n-1)^2/2-n^2(n+1)^2\\ &=\frac12(2n^4-16n^3+11n^2-6n+1). \end{align*}$$

Answer 2

Hint

Consider instead $$\sum_{r=1}^n (r(r+1))^2 x^r$$ and write

$$(r(r+1))^2=r(r-1)(r-2)(r-3)+8r(r-1)(r-2)+14r(r-1)+4r$$ to face derivatives of well know summations.

When done, let $x=-1$.