The coefficient of $x^{n}$ in the expansion of $(2-3 x) /(1-3 x+$ $\left.2 x^{2}\right)$ is

Question

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.

Answer 1

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 $2^n+1$.

Answer 2

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.