Question

The coefficient of $x^{32}$ in the expansion of ${(x^4-\frac{1}{x^3})}^{15}$ is

• A. ${}^{-15}{C}_3$
• B. ${}^{15}{C}_4$
• C. ${}^{-15}{C}_5$
• D. ${}^{15}{C}_2$

Answer 1

The correct option is B ${}^{15}{C}_4$

Given ${(x^4-\frac{1}{x^3})}^{15}$

General term:

$T_{r+1}{=}^{15}{C}_r(x^4{)}^{15-r}.{\left(\frac{-1}{x^3}\right)}^r$

$T_{r+1}{=}^{15}{C}_r{x}^{60-4r}.(-1{)}^r.x^{-3r}$

$T_{r+1}{=}^{15}{C}_r{x}^{60-7r}.(-1{)}^r$

$60-7r=32$

$\therefore r=4$

Coefficient of $x^{32}{=}^{15}{C}_4$