Question

Coefficient of $x^4$ in the expansion of $\frac{1}{(x+1)(x+2)}$ is

• A. $\frac{1}{32}$
• B. $\frac{11}{32}$
• C. $\frac{21}{32}$
• D. $\frac{31}{32}$

Answer 1

The correct option is D $\frac{31}{32}$

Given, $\frac{1}{(x+1)(x+2)}$
$=\frac{x+2-(x+1)}{(x+1)(x+2)}$
$=\frac{1}{x+1}-\frac{1}{x+2}$
$=(1+x{)}^{-1}-(2+x{)}^{-1}$
$=(1+x{)}^{-1}-\frac{1}{2}(1+\frac{x}{2}{)}^{-1}$
Hence coefficient of $x^4$ will be
$\frac{n(n-1)(n-2)(n-3)}{4!}-\frac{1}{2}\frac{n(n-1)(n-2)(n-3)}{4!}\frac{1}{2^4}$
$=\frac{-1(-2)(-3)(-4)}{4!}-\frac{1}{2^5}\frac{-1(-2)(-3)(-4)}{4!}$
$=\frac{4!}{4!}-\frac{1}{32}\frac{4!}{4!}$
$=1-\frac{1}{32}$
$=\frac{31}{32}$