Question

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

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

Answer 1

Answer: (B)

We know by Binomial expansion, that $(x+a{)}^n$
$={}^n{C}_0{x}^n{a}^0+{}^n{C}_1{x}^{n-1}\cdot a+{}^n{C}_2{x}^{n-2}{a}^2$
$+{}^n{C}_3{x}^{n-3}{a}^3+{}^n{C}_4{x}^{n-a}\cdot a^4+...$
$+{}^n{C}_n{x}^0{a}^n$
Given expansion is ${\left(x^4-\frac{1}{x^3}\right)}^{15}$
On comparing we get $n=15,x=x^4$
$a=\left(-\frac{1}{x^3}\right)$
$\therefore {\left(x^4-\frac{1}{x^3}\right)}^{15}$
$={}^{15}{C}_0{\left(x^4\right)}^{15}{\left(-\frac{1}{x^3}\right)}^0$
$+{}^{15}{C}_1{\left(x^4\right)}^{14}\left(-\frac{1}{x^3}\right)+{}^{15}{C}_2{\left(x^4\right)}^{13}{\left(-\frac{1}{x^3}\right)}^2$
$+{}^{15}{C}_3{\left(x^4\right)}^{12}{\left(-\frac{1}{x^3}\right)}^3+{}^{15}{C}_4{\left(x^4\right)}^{11}{\left(-\frac{1}{x^3}\right)}^4+...$
$T_{r+1}={}^{15}{C}_r{\left(x^4\right)}^{15-r}{\left(-\frac{1}{x^3}\right)}^r$
$={}^{-15}{C}_r{x}^{60-7r}$
$\Rightarrow x^{60-7r}=x^{32}$
$\Rightarrow 60-7r=32$
$\Rightarrow 7r=28$
$\Rightarrow r=4$
So, $5th$ term, contains $x^{32}$
$={}^{15}{C}_4{\left(x^4\right)}^{11}{\left(-\frac{1}{x^3}\right)}^4$
$={}^{15}{C}_4{x}^{44}{x}^{-12}={}^{15}{C}_4{x}^{32}{x}^{32}$
$={}^{15}{C}_4$