Autohelmvt

2022-01-24

Prove that:
$\sum _{n=1}^{\mathrm{\infty }}\frac{{n}^{2}\left(n-1\right)}{{2}^{n}}=20$

Troy Sutton

Expert

Note that $\sum _{n}{t}^{n}=\frac{1}{1-t}$ for every $|t|<1$ hence, differentiaiting this twice and three times,
$\sum _{n}n\left(n-1\right){t}^{n-2}=\frac{2}{{\left(1-t\right)}^{3}}$,
$\sum _{n}n\left(n-1\right)\left(n-2\right){t}^{n-3}=\frac{6}{{\left(1-t\right)}^{4}}$
For $t=\frac{1}{2}$, this reads
$\sum _{n}n\left(n-1\right)\frac{1}{{2}^{n-2}}=2\cdot {2}^{3}$
$\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n-3}}=6\cdot {2}^{4}$
which implies
$\sum _{n}n\left(n-1\right)\frac{1}{{2}^{n}}=\frac{1}{4}\cdot 2\cdot {2}^{3}=4$
$\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n}}=\frac{1}{8}\cdot 6\cdot {2}^{4}=12$
Finally,
${n}^{2}\left(n-1\right)=2\cdot n\left(n-1\right)+n\left(n-1\right)\left(n-2\right)$
hence
$\sum _{n}{n}^{2}\left(n-1\right)\frac{1}{{2}^{n}}=2\cdot \sum _{n}n\left(n-1\right)\frac{1}{{2}^{n}}+\sum _{n}n\left(n-1\right)\left(n-2\right)\frac{1}{{2}^{n}}=2\cdot 4+12$
This approach can be made shorter if one notices once and for all that, for every nonnegative k,
$\sum _{n}n\left(n-1\right)\dots \left(n-k\right)\frac{1}{{2}^{n}}=2\cdot \left(k+1\right)!$

egowaffle26ic

Expert

For $|x|<1$ we have that $\sum _{n=0}^{\mathrm{\infty }}{x}^{n}=\frac{1}{1-x}$ hence differentiating both sides we get
$\sum _{n=0}^{\mathrm{\infty }}\left(n+1\right){x}^{n}=\sum _{n=1}^{\mathrm{\infty }}n{x}^{n-1}=\frac{1}{{\left(1-x\right)}^{2}}$
while differentiating once more
$\sum _{n=0}^{\mathrm{\infty }}\left(n+2\right)\left(n+1\right){x}^{n}=\frac{2}{{\left(1-x\right)}^{3}}$
and once again
$\sum _{n=0}^{\mathrm{\infty }}\left(n+3\right)\left(n+2\right)\left(n+1\right){x}^{n}=\frac{6}{{\left(1-x\right)}^{4}}$
Next we express ${n}^{2}\left(n-1\right)$ as a linear combination of $\left(n+3\right)\left(n+2\right)\left(n+1\right),\left(n+2\right)\left(n+1\right),\left(n+1\right)$ and 1:
${n}^{2}\left(n-1\right)=\left(n+3\right)\left(n+2\right)\left(n+1\right)-7\left(n+2\right)\left(n+1\right)+10\left(n+1\right)-2$
and hence
$\sum _{n=0}^{\mathrm{\infty }}{n}^{2}\left(2n-1\right){x}^{n}=\frac{6}{{\left(1-x\right)}^{4}}-\frac{14}{{\left(1-x\right)}^{3}}+\frac{10}{{\left(1-x\right)}^{2}-\frac{2}{1-x}}$
and setting $x=\frac{1}{2}$ we obtain that
$\sum _{n=0}^{\mathrm{\infty }}\frac{{n}^{2}\left(n-1\right)}{{2}^{n}}=6\cdot 16-4\cdot 8+10\cdot 4-4=20$

Do you have a similar question?