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.

Your approach is kind of brute-force, but you can already get the answer using $2(2^{n+1}-1)-3(2^{(n+1)-1}-1) = (4-3)2^n-2+3=(b).$
Note that there’s a $-1$ in the exponent of the term multiplied with $-3$ because to get an $x^n$ term, when the first term is $x^1$, you need the $x^{n-1}$ term.
