This is an exercise in my course notes.
Exercise 3.1 : Find the generating function for the sequence $a_{n}=2^{n}n^{2}$. Hint: Look at the solutions to the generating functions $\displaystyle\frac{1}{(1 - 2x)^{k}}$ and the fact that $n^{2}$ can be written in terms of $\displaystyle\binom{n}{2}$, $n$, and $1$.
I think I have made some progress towards the solution but I haven't gotten all the way. I will present my work below.
I think that the "term formula" for a generating function $\frac{1}{(1 - 2x)^k}$ is $a_n = \binom{n - 1 + k}{n}2^n$.
I figure I'm going to have to pick a value of $k$ to solve this problem. If I let $k = 2$, then the binomial coefficient becomes $\binom{n + 1}{n} = n + 1$. If I let $k = 3$ then the binomial coefficient becomes $\binom{n + 2}{n} = \frac{(n + 2)(n + 1)}{2}$. I'm leaning towards $k = 3$ at the moment but I'm not sure.
Regarding the hint about $\binom{n}{2}, n$ and $1$, I looked at the fact that $\binom{n}{2} = \frac{n^2 - n}{2}$, and I then added $\frac{n^2 + n}{2}$ in order to make it equal $n^2$. Thus I determined that $n^2 = \binom{n}{2} + \binom{n + 1}{2}$.
I thought there might be some binomial coefficient identity that I could take advantage of here, but I didn't see anything relevant at Wikipedia. I think I need to somehow manipulate $\binom{n - 1 + k}{n}$ and $\binom{n}{2} + \binom{n + 1}{2}$ into the same thing, and/or maybe multiply $\frac{1}{(1 - 2x)^k}$ by some constant. I'm not sure what to do, though.
I appreciate any help.
