There is a problem on this site that solves this problem for triangular numbers... However, I wasn't able to apply the same reasonaning for pentagonal numbers.
In general the $kth$ pentagonal number is given by $a_k=(3k^2-k)/2$. Now, I am meant to find the generating function:
$f(x)=a_o+a_1x+a_2x^2+...$, Where the coefficients are the pentagonal numbers.
In the case of the triangular numbers, the solution involved expanding
$1/(1-x)=\sum x^k$ and using some manipulations to get this into the closed form needed. Any advice would be much appreciated. On and ending note, I am wondering if anyone has any resources to direct me to for finding generating functions. The class I am taking is using simple recurrences and sequences, so nothing too difficult.
\begin{eqnarray*} \sum_{i=0}^{\infty}ix^{i}=\frac{x}{(1-x)^2 } \\ \sum_{i=0}^{\infty}i^2x^{i}=\frac{x(1+x)}{(1-x)^3 } \\ \sum_{i=0}^{\infty}\frac{(3i^2-i)}{2}x^{i}=\frac{3}{2}\sum_{i=0}^{\infty}i^2x^{i}-\frac{1}{2}\sum_{i=0}^{\infty}ix^{i}=\frac{2x^2+x}{(1-x)^3 } \\ \end{eqnarray*}