Question

The coefficient of $x^{20}$ in the expansion of ${(1+x^2)}^{40}$ ${(x^2+2+\frac{1}{x^2})}^{-5}$ is

• A. ${}^{30}{\text{C}}_{10}$
• B. ${}^{30}{\text{C}}_{25}$
• C. $1$
• D. none of theses

Answer 1

The correct option is B ${}^{30}{\text{C}}_{25}$

$(1+x^2{)}^{40}(x^2+2+\frac{1}{x^2}{)}^{-5}$
$=(1+x^2{)}^{40}(x+\frac{1}{x}{)}^{-10}$
$=(1+x^2{)}^{40}(1+x^2{)}^{-10}{x}^{10}$
$=x^{10}(1+x^2{)}^{30}$
Writing the general term, we get
$T_{r+1}={}^{30}{C}_r{x}^{2r+10}$
Hence for coefficient of $x^{20}$
$2r+10=20$
$r=5$
Hence the coefficient will be
$T_6={}^{30}{C}_5$
$={}^{30}{C}_{30-5}$
$={}^{30}{C}_{25}$