Infinite sum question

$\sum_{n=1}^{\infty}n^3x^n=\frac{x(x^2+4x+1)}{(1-x)^4}$ for $|x|<1$ . For $x=\frac{1}{3}$ this gives $\frac{33}{8}$ , and the answer is $\boxed{41}$

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

$\displaystyle S(1-\frac{1}{3}) = \frac{1}{3} + \frac{2^3-1^3}{3^{n+1}} + ... = \frac{1}{3}+\sum_{n=1}^{\infty}\frac{(n+1)^3-n^3}{3^{n+1}} =$

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

Following a similar process with $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{3^n}$ results in $\displaystyle\frac{3}{2}( \frac{1}{3} + \sum_{n=1}^{\infty}\frac{2n}{3^{n+1}} + \frac{1}{3^{n+1}}) =$

$\displaystyle\frac{3}{2}( \frac{1}{3}+ \frac{1}{2}+\frac{1}{6} ) = 1.5$

In doing so we see that $\displaystyle\sum_{n=1}^{\infty}\frac{n}{3^n} = \frac{3}{4}$ and

$\displaystyle\sum_{n=1}^{\infty}\frac{1}{3^{n+1}} = \frac{1}{6}$

$\displaystyle\frac{2}{3}S = \frac{1}{3} + \frac{3}{2} + \frac{3}{4} + \frac{1}{6}$

$\displaystyle S = \boxed{\frac{33}{8}}$