How can I find a closed form for this partial sum. $\sum_{n=1}^{k}\frac{n^3}{3^n}$ I am looking for a way to derive the partial sum formula for $ \large \sum_{n=1}^{k}\frac{n^3}{3^n}$. 
I notice that the wolframalpha website evaluates it as $ \frac 1 8  \frac{-4 k^3-18 k^2-36 k+33 (3^k-1)  }{3^k} $  but there is no indication how it derived this.
http://www.wolframalpha.com/input/?i=sum+n%5E3%2F3%5En,n%3D1..k
In general I am looking for a way to find a closed form expression for $ f(x) = \large \sum_{n=1}^{k}x^n n^m$ where $m$ is a positive integer and $x$ is a real number. I am looking at easier cases first to see if a general pattern emerges.
Using the suggestion in the comments, using the geometric sum formula we can differentiate both sides of the expression
$ \large \sum_{n=0}^{k-1} x ^n = \frac{1-x^k}{1-x}$ whence we derive
$ \large \sum_{n=0}^{k-1} nx ^{n-1} = \frac{(k-1)x^k - kx^{k-1}+1}{(1-x)^2}$
Then multiplying both sides by $x$ will give us a closed form expression for $ \sum_{n=0}^{k-1} nx ^{n}$. 
Now we can take the derivative again. 
When you take the derivative of a sum does the upper limit and lower limit change? I did read once that you can increase the lower limit by 1 each time. It looks like the $n=0$ term vanishes here anyway so there is no harm in starting index at $0$.

As an alternative to Ahmed's proposal, you may compute 
$$ T_k(x) = \sum_{n=1}^{k}\frac{e^{nx}}{3^n} $$
then evaluate $T_k'''(0)$, since
$$ T_k'''(x) = \sum_{n=1}^{k}\frac{n^3 e^{nx}}{3^n}. $$
