<s>[INST] Operating with Exponential Generating Function I am having troubles understanding how to operate with generating functions to obtain a final formula. I have been looking to the answer to this question: Exponential generating function and number of balls .

$$\color{red}{\left(\frac{x^2}{2!}+\frac{x^4}{4!}\right)}\color{green}{\frac12\left(e^x+e^{-x}\right)}\color{blue}{e^x}.$$
  This expands to
  $$\frac12\cdot\frac{x^2}{2!} + \frac12\cdot\frac{x^4}{4!} + \sum_{n=2}^\infty \frac1{16}2^nn(n-1)\frac{x^n}{n!} + \sum_{n=4}^\infty\frac1{768}2^nn(n-1)(n-2)(n-3)\frac{x^n}{n!}. $$
  Simplification yields
  $$\frac{x^2}{2!}+3\frac{x^3}{3!}+13\frac{x^4}{4!} + \sum_{n=5}^\infty \frac1{768} 2^nn(n-1)(n^2-5n+54)\frac{x^n}{n!}. $$

What are the steps to obtain this and the simplification? I can't see how to get from one to another
I have tried a different approach, since
$$ e^x + e^{-x} =
\sum_{n=0}^\infty {2x^{2n}\over (2n)!}; \; e^x =
\sum_{n=0}^\infty {x^{n}\over n!}, $$
Multiplying everything, I obtain:
$$
 \frac{1}{2}\frac{x^2}{2!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right) + \frac{1}{2}\frac{x^4}{4!}\left(\sum_{n=0}^\infty\frac{2x^{2n}}{(2n)!}\right)\cdot \left(\sum_{n=0}^\infty \frac{x^n}{n!}\right)
$$
Removing the 2 of the numerator of the fractions and simplifying $x^{2n}/2n!$ for $x^{n}/n!$;  and if I understand multiplication correctly, I should have:
$$
\sum_{k=0}^{n}\binom{n}{k}\frac{x^n}{n!}
$$
as a result of both the multiplications, but doesn't look like any form of the right answer.
PS: Decided to make another question instead of commenting the answer because it is 2 years old already.
 [/INST] The term with $\frac{1}{16}$ arises as follows:
\begin{align*}
  \frac{x^2}{2}\cdot\frac{1}{2}e^{2x}
     &= \frac{1}{4}x^2\sum_{n=0}^\infty \frac{2^nx^n}{n!} \\
     &= \frac{1}{4}\sum_{n=0}^\infty \frac{2^nx^{n+2}}{n!} \\
     &= \frac{1}{16}\sum_{n=0}^\infty \frac{2^{n+2}x^{n+2}}{n!} \\
     &= \frac{1}{16}\sum_{n=2}^\infty \frac{2^nx^n}{(n-2)!} \\
     &= \frac{1}{16}\sum_{n=2}^\infty n(n-1)\frac{2^nx^n}{n!}.
\end{align*}
The other term arises similarly, starting from $\frac{x^4}{24}\cdot\frac{1}{2}e^{2x}$.
</s>