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 $\displaystyle \sum_{n=1}^\infty \frac{n^3}{2^n}$ 

As hinted by @AlexandreHalm,
 $$\sum_{k=0}^\infty x^k=\frac1{1-x}.$$
Then deriving on $x$ and multiplying by $x$,
$$\sum_{k=0}^\infty kx^k=\frac x{(1-x)^2},$$
 $$\sum_{k=0}^\infty k^2x^k=\frac{2x^2}{(1-x)^3}+\frac x{(1-x)^2},$$
 $$\sum_{k=0}^\infty k^3x^k=\frac{6x^3}{(1-x)^4}+\frac{6x^2}{(1-x)^3}+\frac x{(1-x)^2}.$$
Evaluating at $x=\frac12$, 
$$\sum_{k=0}^\infty\frac{k^3}{2^k}-\sum_{k=0}^\infty\frac{k^2}{2^k}=26-6=20.$$
