Question

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

Answer 1

The general term in the given expansion is given by
$T_{r+1}=(-1{)}^r{\times }^{15}{C}_r\times (x^4{)}^{(15-r)}\times {\left(\frac{1}{x^3}\right)}^r\Rightarrow T_{r+1}=(-1{)}^r{\times }^{15}{C}_r\times x^{(60-7r)}\text{Putting}60-7r=32,\text{we get}7r=28\Rightarrow r=4\Rightarrow r+1=5\text{Now},T_5=T_{(4+1)}=(-1{)}^4{\times }^{15}{C}_4\times x^{(60-28)}{=}^{15}{C}_4\times x^{32}.\therefore \text{Coefficent of}{x}^{32}{=}^{15}{C}_4=\left(\frac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}\right)=1365.\text{let}{T}_{s+1}\text{be the term containing}{x}^{-17}\text{Then,}{T}_{s+1}=(-1{)}^s{\times }^{15}{c}_s\times x^{(60-7s)}\text{Putting 60 - 7x = - 17, we get s = 11 and therefore , s + 1 = 12}\text{Thus, the 12th term contains}{x}^{-17}\text{Now,}{T}_{12}=T_{(11+1)}=(-1{)}^{11}{\times }^{15}{C}_{11}\times x^{(60-77)}={-}^{15}{C}_4\times x^{-14}\therefore \text{Coefficient of}{x}^{-17}={-}^{15}{C}_4=-\left(\frac{15\times 14\times 13\times 12}{4\times 3\times 2\times 1}\right)=-1365.$
So, the coefficient of $x^{17}$ in the given expansion is -1365
hence, the coefficient of $x^{32}$ is 1365 and that of $x^{-17}$ is - 1365