Question

The coefficient of $x^n$ is $\frac{x-4}{x^2-5x+6}$ is

• A. $\frac{1}{3^{n+1}}-\frac{1}{2^n}$
• B. $\frac{1}{3^{n+1}}+\frac{1}{2^n}$
• C. $\frac{1}{5^{n+1}}+\frac{1}{2^n}$
• D. $\frac{1}{5^{n+1}}+\frac{1}{3^n}$

Answer 1

The correct option is A $\frac{1}{3^{n+1}}-\frac{1}{2^n}$

$\frac{x-y}{x^2-5x+6}=\frac{(x-y)}{(x-3)(x-2)}$
$\frac{x-4}{(x-3)(x-2)}=\frac{A}{x-3}+\frac{B}{x}-2$
$B=+2;A=-1$
$=\frac{-1}{x-3}+\frac{2}{x-2}$
And co-efficient of $x^{n}is(x-a{)}^{-1}$ is $\frac{-1}{a^{n+1}}$
So co-efficient of $x^n$ in $-1/x-3$ is $1/3^{n+1}$ --(1)
Co-efficient of $x^n$ in $\frac{1}{x-2}is-1/2^{n+1}$
i.e $\frac{2}{x-2}=-1/2^n--\left(2\right)$
From (1) & (2) it is $1/3^{n+1}-1/2^n$