<s>[INST] The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is The coefficient of $x^{n}$ in the expansion of $\frac{(2-3 x)}{(1-3 x+\left.2 x^{2}\right)}$ is
$(a) \quad(-3)^{n}-(2)^{n / 2-1}$
(b) $2^{n}+1$
$(c) 3(2)^{n / 2-1}-2(3)^{n}$
(d) None of the foregoing numbers.
Now, $1-3 x+2 x^{2}=(1-x)(1-2 x)$
Now, $(1-x)^{-1}$
$=1+x+x^{2}+x^{3}+\ldots$
Now, $(1-2 x)^{-1}$
$=1+2 x+(2 x)^{2}+(2 x)^{3}+\ldots \ldots$
Coefficient of $x^{n}$ in $(1-x)^{-1}(1-2 x)^{-1}=2^{n}+2^{n-1}+2^{n-2}+\ldots .+2+1=$
$2^{n+1}-1$
What to do next??
Any shortcut or objective approach for this type of problems would be highly appreciated.
 [/INST] You're on the right track. Consider splitting the fraction up using partial fractions.
\begin{align*}
\frac{2-3x}{1-3x+2x^2} &= \frac{2-3x}{(1-2x)(1-x)}\\ &=\frac{1}{1-2x} + \frac{1}{1-x}\\&=\sum_{n=0}^\infty 2^nx^n + \sum_{n=0}^\infty x^n\\&=\sum_{n=0}^\infty \left (2^n+1\right )x^n
\end{align*}
Hence, the coefficient of $x^n$ is $2^n+1$.
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