Finding the coefficient in the closed form of the generating function I try to solve the recursion $a_n=5a_{n-1}+5^n$ with $a_0=1$ with generating function, but I could not find the coefficient of $x^n$ in the closed form
\begin{eqnarray*}
g(x)&=&a_0+\sum^n_1a_nx^n\\
g(x)&=&1+\sum^n_1(5a_{n-1}+5^n)x^n\\
g(x)&=&1+5\sum^n_1a_{n-1}x^n + \sum^n_15^nx^n\\
g(x)&=&1+5x\sum^n_0a_{n-1}x^{n-1} + \sum^n_1(5x)^n\\
g(x)&=&1+5xg(x)+\frac{1}{1-5x}-1\\
g(x)(1-5x)&=&\frac{1}{1-5x}\\
g(x)&=&\frac{1}{(1-5x)^2}
\end{eqnarray*}
I tried the partial fraction and wish to find something in the form $\frac{A}{1-5x}+\frac{B}{1-5x}$, but it did not work out because the method of partial fraction required this in form of $\frac{A}{1-5x}+\frac{B}{(1-5x)^2}$, so it seems like partial fraction does not help. Can someone help.

You get the generating function:
$$
g(z) = \frac{1}{(1 - 5 z)^2}
     = \sum_{n \ge 0} \binom{-2}{n} (-1)^n 5^n z^n
     = \sum_{n \ge 0} \binom{n + 2 - 1}{2 - 1} 5^n z^n
     = \sum_{n \ge 0} (n + 1) \cdot 5^n z^n
$$
so that:
$$
a_n = (n + 1) \cdot 5^n
$$
