<s>[INST] Prove that $\sum\limits_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$ This sum  $\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} $showed up as I was computing the expected value of a random variable.
My calculator tells me that  $\,\,\displaystyle \sum_{n=1}^\infty \frac{n^2(n-1)}{2^n} = 20$.
How can I show that?
I know how to find the value of the sum $\,\displaystyle \sum_{n=1}^\infty \frac{n^2}{2^n},\,$ but I can't deal with $\displaystyle \sum_{n=1}^\infty \frac{n^3}{2^n}$ 
 [/INST] For $\lvert x\rvert<1$ we have that
$\displaystyle\sum_{n=0}^\infty x^n=\frac{1}{1-x},$
hence differentiating both sides we get
$$
\sum_{n=0}^\infty (n+1)x^{n}=\sum_{n=1}^\infty nx^{n-1}=\frac{1}{(1-x)^2},
$$
while differentiating once more
$$
\sum_{n=0}^\infty (n+2)(n+1)x^{n}=\frac{2}{(1-x)^3},
$$
and once again
$$
\sum_{n=0}^\infty (n+3)(n+2)(n+1)x^{n}=\frac{6}{(1-x)^4}.
$$
Next, we express $n^2(n-1)$ as a linear combination of
$(n+3)(n+2)(n+1)$, $(n+2)(n+1)$, $(n+1)$ and $1$:
$$
n^2(n-1)=(n+3)(n+2)(n+1)-7(n+2)(n+1)+10(n+1)-2,
$$
and hence
$$
\sum_{n=0}^\infty n^2(n-1)x^n=\frac{6}{(1-x)^4}-\frac{14}{(1-x)^3}+\frac{10}{(1-x)^2}-\frac{2}{1-x},
$$
and setting $x=1/2$ we obtain that
$$
\sum_{n=0}^\infty \frac{n^2(n-1)}{2^n}=6\cdot 16-4\cdot 8+10\cdot 4-4=20.
$$
</s>