enumerative combinatorics and geometric series So, I'm going to take an enumerative combinatorics class this upcoming semester.  I began reading about it and came across and interesting example, but I am not sure how they arrive at their final answer. The example is in in the image I included.  I don't know how they determined the equations for $c_n$.  Any help would be greatly appreciated.
 A: Are you asking about the following statement?
$\displaystyle \sum_{n \geq 0} \alpha^n x^n = \frac{1}{1 - \alpha x}$
This is nothing but the sum of a geometric series. For a derivation, see 
Geometric series
A: This is standard
manipulation of power series,
and you will have to
become comfortable
with doing this.
You want to get the power series for
$\left(\sum_{n=0}^{\infty} a^nx^n\right)(1-ax)
$.
In this case,
you separate the two sums,
shift an index of summation,
and recombine the two series.
Here are the details.
$\begin{array}\\
P(x)
&=\left(\sum_{n=0}^{\infty} a^nx^n\right)(1-ax)\\
&=\left(\sum_{n=0}^{\infty} a^nx^n\right)-ax\left(\sum_{n=0}^{\infty} a^nx^n\right)
\qquad\text{distribute the sum}\\
&=\left(\sum_{n=0}^{\infty} a^nx^n\right)-\left(\sum_{n=0}^{\infty} a^{n+1}x^{n+1}\right)
\qquad\text{bring }ax \text{ into the sum}\\
&=\left(\sum_{n=0}^{\infty} a^nx^n\right)-\left(\sum_{n=1}^{\infty} a^{n}x^{n}\right)
\qquad\text{shift the index of summation}\\
&=1+\left(\sum_{n=1}^{\infty} a^nx^n\right)-\left(\sum_{n=1}^{\infty} a^{n}x^{n}\right)
\qquad\text{split the first term so the indices match}\\
&=1
\qquad\text{like magic, the two series cancel each other out}\\
\end{array}
$
