Question

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

• A. $\frac{3^{n+1}-1}{{2.3}^{n+1}}$
• B. $\frac{3^{n+1}-1}{3^{n+1}}$
• C. $2\left(\frac{3^{n+1}-1}{3^{n+1}}\right)$
• D. None of these

Answer 1

The correct option is A $\frac{3^{n+1}-1}{{2.3}^{n+1}}$

$\frac{1}{(1-x)(3-x)}=\frac{1}{3}.\frac{1}{(1-x)(1-\frac{x}{3})}$

$\frac{1}{3}{(1-x)}^{-1}{(1-\frac{x}{3})}^{-1}$

$=\frac{1}{3}(1+x+x^2+.....+x^n+...)(1+\frac{x}{3}+\frac{x^2}{3^2}+....)$

Coefficient of $x^n$ in $\frac{1}{(1-x)(3-x)}$

$=\frac{1}{3}$( coefficient of $1$ in ${(1-x)}^{-1}\times$ coefficient of $x^n$ in ${(1-\frac{x}{3})}^{-1}+$ coefficient of $x$ in ${(1-x)}^{-1}\times$ coefficeint of $x^{n-1}$ in ${(1-\frac{x}{3})}^{-1}+....+$ coefficeint of $x^n$ in ${(1-x)}^{-1}\times$ coefficient of$1$ in ${(1-\frac{x}{3})}^{-1})$

$=\frac{1}{3}(1\times \frac{1}{3^n}+1\times \frac{1}{3^{n-1}}+...+1)$

$=\frac{1}{3}\times 1\times \frac{1-\frac{1}{3^{n+1}}}{1-\frac{1}{3}}=\frac{3^{n+1}-1}{2.3^{n+1}}$