Question

The coefficient of $x^5$ in $(1+2x+3x^2+....{)}^{-7/2}$ is

• A. 15
• B. 21
• C. 12
• D. 30

Answer 1

Answer: (B)

$1+2x+3x^2+....={\left(1+x\right)}^{-2}$
$\Rightarrow {\left(1+2x+3x^2+....\right)}^{-3/2}={\left\{{\left(1+x\right)}^{-2}\right\}}^{-7/2}={\left(1+x\right)}^7$
$\therefore$ The coefficient of $x^5$ in ${\left(1+2x+3x^2+........\right)}^{-7/2}$
= Coefficient of $x^5$ in ${\left(1+x\right)}^7$
${=}^7{C}_5=21$