<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] Note that $\sum\limits_nt^n=\frac1{1-t}$ for every $|t|\lt1$ hence, differentiating this twice and three times, $$\sum_nn(n-1)t^{n-2}=\frac2{(1-t)^3},\qquad\sum_nn(n-1)(n-2)t^{n-3}=\frac6{(1-t)^4}.$$ For $t=\frac12$, this reads $$\sum_nn(n-1)\frac1{2^{n-2}}=2\cdot2^3,\qquad\sum_nn(n-1)(n-2)\frac1{2^{n-3}}=6\cdot2^4,$$ which implies $$\sum_nn(n-1)\frac1{2^{n}}=\frac14\cdot2\cdot2^3=\color{red}{4},\qquad\sum_nn(n-1)(n-2)\frac1{2^{n}}=\frac18\cdot6\cdot2^4=\color{blue}{12}.$$ Finally, $$n^2(n-1)=\color{green}{2}\cdot n(n-1)+n(n-1)(n-2),$$ hence $$\sum_nn^2(n-1)\frac1{2^{n}}=\color{green}{2}\cdot\sum_nn(n-1)\frac1{2^{n}}+\sum_nn(n-1)(n-2)\frac1{2^{n}}=\color{green}{2}\cdot\color{red}{4}+\color{blue}{12}.$$ This approach can be made shorter if one notices once and for all that, for every nonnegative $k$, $$\sum_nn(n-1)\cdots(n-k)\frac1{2^{n}}=2\cdot(k+1)!.$$
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