An unusual power value in power series Let the power series to be defined as :
$$\sum\limits_{n=0}^\infty a_nx^n$$
How to determine $a_{n}$ in such cases?
$$\begin{eqnarray}&a)&\quad \sum\limits_{n=0}^\infty \frac {n^{n}} {n!} x^{3n}\\ \\
&b)&\quad \sum\limits_{n=0}^\infty \frac {n}{2^{n}(x+2)^{2n}}\end{eqnarray}$$
 A: Just read off the coefficients. For (a), for instance, you have
$$a_n=\begin{cases}
\frac{k^k}{k!},&\text{if }n=3k\\\\
0,&\text{otherwise}\;.
\end{cases}$$
A: The two problems are so different from each other that it leaves me wondering if they're even supposed to be part of the same exercise, or if they may be mis-transcribed in some fashion.  That said, here's one approach to solving (b):
a) Expand out $\displaystyle\frac{1}{(x+2)^{2n}}$ into a power series in $x$ using the Binomial Series :
$$\begin{eqnarray}\frac{1}{(x+2)^{2n}} &=& \frac{1}{\left(2(\frac{x}{2}+1)\right)^{2n}} \\
&=&\frac{1}{2^{2n}}\frac{1}{\left(\frac{x}{2}+1\right)^{2n}} \\
&=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\binom{i+2n-1}{i}\left(-\frac{x}{2}\right)^i \\
&=&\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i
\end{eqnarray}$$
b) Plug this expansion into your original power series:
$$\begin{eqnarray}\sum_{n=0}^\infty\frac{n}{2^n\left(x+2\right)^{2n}} &=&\sum_{n=0}^\infty\frac{n}{2^n}\frac{1}{\left(x+2\right)^{2n}}\\
&=&\sum_{n=0}^\infty\frac{n}{2^n}\left(\frac{1}{2^{2n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\
&=&\sum_{n=0}^\infty\left(\frac{n}{2^{3n}}\sum_{i=0}^\infty\frac{(-1)^i}{2^i}\binom{i+2n-1}{i}x^i\right) \\
\end{eqnarray}$$
c) Exchange the orders of summation so that the sum over $i$ is your outer summation; you'll be left with something that looks like $\displaystyle\sum_{i=0}^\infty\left(\sum_{n=0}^\infty b_{n,i}\right)x^i$ where $b_{n,i}$ is some complicated expression in $n$ and $i$.  Your coefficients $a_i$ are then just $\displaystyle\sum_{n=0}^\infty b_{n,i}$.
