Question

Find the coefficient of $x^{32}$ and $x^{-17}$ in ${(x^4-\frac{1}{x^3})}^{15}$.

Answer 1

General term $T_{r+1}$ in the expansion of $(x+y{)}^n$ is given by

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

Using above result, we have
$T_{r+1}=(-1{)}^r{}^{15}{C}_r{\left(x^4\right)}^{15-r}{\left(\frac{1}{x^3}\right)}^r$
$={(-1)}^r\cdot {}^{15}{C}_r\cdot x^{60-4r-3r}$$={(-1)}^r\cdot {}^{15}{C}_r\cdot x^{60-7r}$
For $x^{32},60-7r=32$
i.e. $7r=28$
i.e. $r=4$
$\therefore$ Coefficient of $x^{32}$ is ${}^{15}{C}_4=1365$
For $x^{-17},60-7r=-17$
$7r=77$
$r=11$
Coefficient of $x^{-17}$ is ${(-1)}^{11}{\times }^{15}{C}_{11}$ $=-1365$