Question

The coefficient of $x^{-17}$ in the expansion of ${\left(x^4-\frac{1}{x^3}\right)}^{15}$ is
(a) 1365
(b) −1365
(c) 3003
(d) −3003

Answer 1

(b) −1365

$$\begin{gathered} \mathrm{Suppose}\mathrm{the}(r+1)\mathrm{th}\mathrm{term}\mathrm{in}\mathrm{the}\mathrm{given}\mathrm{expansion}\mathrm{contains}\mathrm{the}\mathrm{coefficient}\mathrm{of}{x}^{-17}. \\ \mathrm{Then},\mathrm{we}\mathrm{have}: \\ {T}_{r+1}={}^{15}C_r(x^4{)}^{15-r}{\left(\frac{-1}{x^3}\right)}^r \\ =(-1{)}^r{}^{15}C_r{x}^{60-4r-3r} \\ \mathrm{For}\mathrm{this}\mathrm{term}\mathrm{to}\mathrm{contain}{x}^{-17},\mathrm{we}\mathrm{must}\mathrm{have}: \\ 60-7r=-17 \\ \Rightarrow 7r=77 \\ \Rightarrow r=11 \\ \therefore \mathrm{Required}\mathrm{coefficient}=(-1{)}^{11}{}^{15}C_{11}=-\frac{15\times 14\times 13\times 12}{4\times 3\times 2}=-1365 \end{gathered}$$