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.

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 $2^n+1$.
