Question

The coefficient of $x^n$ in the expansion of $\frac{1}{(1-x)(1-2x)(1-3x)}$ is

• A. $\frac{1}{2}(2^{n+2}-3^{n+3}+1)$
• B. $\frac{1}{2}(3^{n+2}-2^{n+3}+1)$
• C. $\frac{1}{2}(2^{n+3}-3^{n+2}+1)$
• D. $noneofthese$

Answer 1

Answer: (B)

$\frac{1}{2}(3^{n+2}-2^{n+3}+1)$

We have

$\frac{1}{\left(1-x\right)\left(1-2x\right)\left(1-3x\right)}$

$=\frac{1}{2\left(1-x\right)}-\frac{4}{1-2x}+\frac{9}{2\left(1-3x\right)}$

$=\frac{1}{2}{\left(1-x\right)}^{-1}-4{\left(1-2x\right)}^{-1}+\frac{9}{2}{\left(1-3x\right)}^{-1}$

$=\frac{1}{2}\left(1+x+x^2+.....+x^n\right)-4\left(1+2x+{\left(2x\right)}^2+.....+{\left(2x\right)}^n\right)+\frac{9}{2}\left(1+3x+{\left(3x\right)}^2+....+{\left(3x\right)}^n\right)$

Coefficient of $x^n$, we get

$=\frac{1}{2}\left[1-{8.2}^n+{9.3}^n\right]$

$=\frac{1}{2}\left[1-2^{n+3}+3^{n+2}\right]$

$=\frac{1}{2}\left[3^{n+2}-2^{n+3}+1\right]$

Hence, the option $\left(B\right)$ is correct answer.