The title basically says it all. If $2n$ was just $n$, I think it's relatively simple:
$\sum_{n=0}^{\infty} {n+E-1 \choose n} x^{n} = \frac{1}{(1-x)^E}$.
But what if you want every other term, i.e.
$\sum_{n=0}^{\infty} {2n+E-1 \choose 2n} x^{2n}$
Can anyone think of a simple expression for that?
We have \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{n+E-1}{n} x^n =\frac{1}{(1-x)^E}. \end{eqnarray*} Break this into its odd & even powers of $x$ \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n+E-1}{2n} x^{2n} +\sum_{n=0}^{\infty} \binom{2n+E}{2n+1} x^{2n+1} =\frac{1}{(1-x)^E}=\frac{(1+x)^E}{(1-x^2)^E}. \end{eqnarray*} Now binomially expand the numerator and collect the even powers of $x$ and we have \begin{eqnarray*} \sum_{n=0}^{\infty} \binom{2n+E-1}{2n} x^{2n} =\frac{1}{(1-x^2)^E} \sum_{i=0}^{ \lfloor \frac{E}{2} \rfloor} \binom{E}{2i} x^{2i}. \end{eqnarray*}