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 this post is intended to justify the answer of @Dr.Sonnhard Graubner, I make this CW.
The starting points are that, defining $S_k(m)=\sum_{n=1}^m\frac{n^k}{2^n},$ we can easily compute $S_0(m)$ and $S_k(1),$ and that we can use recursive relations to obtain $S_3(m)$ from $S_0(m)$ and $S_3(1).$
Thus we divide the post into four steps:
I. $k=0$
$S_0(m)=\sum_{n=1}^m\frac{1}{2^n}=1-\frac{1}{2^m}$ (Geometric series)
II. $k=1$
Since $S_1(m)-\frac{1}{2}=\frac{1}{2}(\sum_{n=1}^{m-1}\frac{n+1}{2^n})=\frac{1}{2}S_1(m-1)+\frac{1}{2}S_0(m-1),$ we find that $S_1(m)=\frac{1}{2}S_1(m-1)+1-2^{-m}=\frac{1}{2}(\frac{1}{2}S_1(m-2)+1-2^{-m+1})+1-2^{-m}=\cdots=\frac{1}{2^{m-1}}S_1(1)+\sum_{n=0}^{m-2}\frac{1}{2^{n}}(1-\frac{1}{2^{m-n}})=\frac{1}{2^m}+\frac{1-(1/2)^{m-1}}{1/2}-\frac{(m-1)}{2^m}=2-\frac{1}{2^m}(m+2).$
III. $k=2$
By similar token, one finds the recursive relation $S_2(m)=\frac{1}{2}S_2(m-1)+3-\frac{2m+3}{2^m}.$ Then we deduce the formular for $S_2(m):$
$$S_2(m)=\frac{1}{2^m}+\sum_{n=0}^{m-2}\frac{1}{2^n}(3-\frac{2(m-n)+3}{2^{m-n}})=\frac{1}{2^m}+6-\frac{12}{2^m}-\frac{1}{2^m}(m^2+4m-5)$$$$=6-\frac{1}{2^m}(m^2+4m+6).$$
IV. $k=3$
Again we have the resursive relation $S_3(m)=\frac{1}{2}S_3(m)+13-\frac{1}{2^m}(3m^2+9m+13),$ from which we deduce the formula:
$$S_3(m)=\frac{1}{2^{m-1}}S_3(1)+\sum_{n=0}^{m-2}\frac{1}{2^n}(13-\frac{3(m-n)^2+9(m-n)+13}{2^{m-n}})$$
$$=\cdots=\frac{1}{2^m}+26-\frac{52}{2^m}-\frac{m^3+6m^2+18m-25}{2^m}=26-\frac{m^3+6m^2+18m+26}{2^m}.$$
Here the dots mean a process of simplification.
Hence we see that $\lim_{m\rightarrow\infty}S_3(m)-S_2(m)=26-6=20.$
So I am wondering where do those $(1/2)^{m+1}$ terms in the answer by @Dr.Sonnhard Graubner appear.  

 Finally I would like to say something, in reply to @Did: Indeed no simple functional equation holds for $S_k(m),$ but one doesn't expect there to be one at all: instead one looks for a recursive relation, to which one is used in dealing with discrete relations, such as finding the partial sums.  

Hope this helps still.
