Question

Find the coefficient of $x^9$ in the expansion of ${(11x^2+\frac{3}{x^5})}^{15}$.

• A. ${}^{15}{C}_{12}{11}^3{3}^{12}$
• B. ${}^{15}{C}_3{11}^{10}{3}^5$
• C. ${}^{15}{C}_3{3}^2{11}^3$
• D. ${}^{15}{C}_{12}{11}^{12}{3}^3$

Answer 1

The correct option is D ${}^{15}{C}_{12}{11}^{12}{3}^3$

Given ${(11x^2+\frac{3}{x^5})}^{15}$

General term:-

$T_{r+1}={}^n{C}_r{x}^{n-r}{y}^r$

$T_{r+1}={}^{15}{C}_r(11r^2{)}^{15-r}{\left(\frac{3}{x^5}\right)}^r$

$T_{r+1}={}^{15}{C}_r\left(11{)}^{15-r}\right(x{)}^{30-2r}\frac{(3{)}^r}{x^{5r}}$

$T_{r+1}={}^{15}{C}_r\left(11{)}^{15-r}\right(x{)}^{30-7r}(3{)}^r$

We need to find coefficient of $x^9$

so, $30-7r=9$

$21=7r$

$r=3$

$T_4={}^{15}{C}_3\left(11{)}^{12}\right(3{)}^3$

$T_4={}^{15}{C}_{12}{11}^{12}{3}^3$ ............ ${}^n{C}_r={}^n{C}_{n-r}$

Hence correct option is D.