Find generating function of $\frac{5-3x}{2-3x+x^2}$.
My attempt: $\frac{5-3x}{2-3x+x^2} = \frac{1}{2} \cdot (5-3x) \cdot \frac{1}{1-x/2} \cdot \frac{1}{1-x}$. Here you use geometric series, but I don't know, how to find the coefficient before $x^n$ in this form.
Any ideas?
On
Another method with your method: $$ \begin{align} \frac{5-3x}{2-3x+x^2}=&\frac{1}{2}\left( 5-3x \right) \frac{1}{1-x}\frac{1}{1-\frac{x}{2}} \\ =&\left( 5-3x \right) \cdot \sum_{n=0}^{+\infty}{x^n}\cdot \frac{1}{2}\sum_{n=0}^{+\infty}{\left( \frac{x}{2} \right) ^n} \\ =&\frac{5-3x}{2}\sum_{n=0}^{+\infty}{\sum_{m=0}^{+\infty}{x^m\left( \frac{x}{2} \right) ^n}} \\ =&\frac{5-3x}{2}\sum_{n=0}^{+\infty}{\sum_{m=0}^{+\infty}{\frac{x^{m+n}}{2^n}}} \\ =&\frac{5-3x}{2}\sum_{k=0}^{+\infty}{\sum_{m+n=k}^{+\infty}{\frac{x^k}{2^n}}} \\ =&\frac{5-3x}{2}\sum_{k=0}^{+\infty}{\sum_{n=0}^k{\frac{1}{2^n}x^k}} \\ =&\frac{5-3x}{2}\sum_{k=0}^{+\infty}{\frac{1-\left( \frac{1}{2} \right) ^{k+1}}{1-\frac{1}{2}}x^k} \\ =&\left( 5-3x \right) \sum_{k=0}^{+\infty}{\left( 1-\left( \frac{1}{2} \right) ^{k+1} \right) x^k} \\ =&\sum_{k=0}^{+\infty}{\left( 5\left( 1-\left( \frac{1}{2} \right) ^{k+1} \right) -3\left( 1-\left( \frac{1}{2} \right) ^k \right) \right) x^k} \\ =&\sum_{k=0}^{+\infty}{\left( 2+\frac{1}{2^{k+1}} \right) x^k} \end{align} $$
$$ \begin{align} \frac{5-3x}{2-3x+x^2}=&\frac{5-3x}{\left( x-1 \right) \left( x-2 \right)} \\ =&\frac{2}{1-x}+\frac{\frac{1}{2}}{1-\frac{x}{2}} \\ =&2\sum_{n=0}^{+\infty}{x^n}+\frac{1}{2}\sum_{n=0}^{+\infty}{\left( \frac{x}{2} \right) ^n} \\ =&\sum_{n=0}^{+\infty}{\left( 2+\frac{1}{2^{n+1}} \right) x^n} \end{align} $$