Finding the coefficient in the closed form of the generating function

Question

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.

Answer 1

(I haven't checked your algebra, but it's plausible. Check you get the right answer at the end by working out a few terms.)

The ingredient you're missing is a series expansion of $(1+u)^{-2}$, which is given by a binomial series, http://en.m.wikipedia.org/wiki/Binomial_series

You get $$\sum_0^\infty x^n \frac{(-2)(-3)(-4)(-5)\cdots (-n-1)}{(1)(2)(3)(4)\cdots (n)}$$

and you can simplify the fraction a lot! This should be enough of hint.