Assume that 
has a power-series representation in terms of powers of 
, verify that it has the form
\[ \frac{x}{(1-x)(1-x^2)} = \frac{1}{2} \sum_{n=1}^{\infty} \left( n + \frac{1 – (-1)^n}{2} \right)x^n
valid for all real such that 
.
First, we rewrite the given expression in a form we can deal with,
![\begin{align*} \frac{x}{(1-x)(1-x^2)} &= \frac{x}{(1-x)(1-x)(1+x)} \\[9pt] &= \frac{x}{(1-x)^2(1-x)} \\[9pt] &= \frac{1}{2} \frac{1}{(1-x)^2} - \frac{1}{4} \frac{1}{1-x} - \frac{1}{4} \frac{1}{1+x} &(\text{partial fractions}). \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-835cc624cb94bc1d21bde684df2b2d1e_l3.png "Rendered by QuickLaTeX.com")
Then, we use the geometric series formula, and we call from a previous exercise (Section 11.13, Exercise #3) that 
, so we have
![\begin{align*} \frac{x}{(1-x)(1-x^2)} &= \frac{1}{2} \frac{1}{(1-x)^2} - \frac{1}{4} \frac{1}{1-x} - \frac{1}{4} \frac{1}{1+x} \\[9pt] &= \frac{1}{2} \sum_{n=0}^{\infty} nx^{n-1} - \frac{1}{4} \sum_{n=0}^{\infty} x^n - \frac{1}{4} \sum_{n=0}^{\infty} (-1)^n x^n \\[9pt] &= \frac{1}{2} \left( \sum_{n=0}^{\infty} nx^n - \frac{1}{2} \sum_{n=0}^{\infty} (1 - (-1)^n)x^n \right) \\[9pt] &= \frac{1}{2} \left( \sum_{n=1}^{\infty} nx^n - \frac{1}{2} \sum_{n=1}^{\infty} (1 - (-1)^n) x^n \right) \\[9pt] &= \frac{1}{2} \sum_{n=1}^{\infty} \left( n - \frac{1 - (-1)^n}{2} \right) x^n. \end{align*}](https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-3f4dff815be6eea0d263274e00517f78_l3.png "Rendered by QuickLaTeX.com")
This valid for all real since that was where the geometric series formulas we used were valid.