Question

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

• A. 1
• B. -1
• C. 2
• D. 0

Answer 1

The correct option is D 0

Using Binomial theorem,

${(x^4-\frac{1}{x^3})}^5=\sum _{n=0}^{15}{}^{15}{C}_n(x^4{)}^{15-n}{\left(\frac{-1}{x^3}\right)}^n$

$=\sum _{n=0}^{15}{}^{15}{C}_n{x}^{60-4n}\frac{(-1{)}^n}{x^{3n}}$

$=\sum _{n=0}^{15}(-1{)}^n{}^{15}{C}_n{x}^{60-4n-3n}$

$=\sum _{n=0}^{15}(-1{)}^n{}^{15}{C}_n{c}^{60-7n}$

$T_{r+1}=(-1{)}^n{}^{15}{C}_r{C}^{60-7r}$ $[60-7r=32,7r=60-32,r=\frac{28}{7}=4]$

Coefficient of $x^{32}=(-1{)}^4{}^{15}{C}_4$

$={}^{15}{C}_4$

Coefficient of $x^{-17}=(-1{)}^{11}{}^{15}{C}_{11}$ $[60-7r=-17,\Rightarrow 7r=60+17,\Rightarrow r=\frac{77}{7}=11,$ and using ${}^n{C}_r={}^n{C}_n-r]$

$={-}^{15}{C}_4$

Sum of coefficient of $x^{32}$ and $x^{-17}={}^{15}{C}_4-{}^{15}{C}_4$

$=0$.