Using generating functions, Find a closed formula to next expression: $\sum_{k=0}^m{k(k+2)}$

Question

Using generating functions, Find a closed formula to next expression:

$\sum_{k=0}^m{k(k+2)}$

If i use calculus power series rules, The question is fairly simple. But how can i find the proper relation with generating functions?

Answer 1

First note that

$$\sum_{k=0}^nk(k+2)x^k=\sum_{k=0}^n(k+1)^2x^k-\sum_{k=0}^nx^k\;.$$

The last summation is a geometric series, so you know a closed form for it. Next, check that

$$\sum_{k=0}^n(k+1)^2x^k=\sum_{k=1}^{n+1}k^2x^{k-1}=\left(\sum_{k=1}^{n+1}kx^k\right)'=\left(\sum_{k=0}^{n+1}kx^k\right)'\;.$$

If you can find a closed form for $\displaystyle\sum_{k=0}^{n+1}kx^k$, you can differentiate it to get one for $\displaystyle\sum_{k=0}^n(k+1)^2x^k$. And

$$\sum_{k=0}^{n+1}kx^k=\sum_{k=1}^{n+1}kx^k=x\sum_{k=1}^{n+1}kx^{k-1}=x\left(\sum_{k=1}^{n+1}x^k\right)'=x\left(\sum_{k=0}^{n+1}x^k\right)'\;,$$

for which you can easily find a closed form.

Answer 2

Hint: Consider the function $$ u_m(x)=\sum_{k=0}^mx^k=\frac{1-x^{m+1}}{1-x}, $$ then $$ u'_m(x)=\sum_{k=0}^mkx^{k-1},\quad u''_m(x)=\sum_{k=0}^mk(k-1)x^{k-2}, $$ hence $$ \sum_{k=0}^mk(k+2)=u''_m(1)+3u'_m(1). $$ Can you compute $u'_m(1)$ and $u''_m(1)$?