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 $n=15, x=x^{4}$
$a=\left(-\frac{1}{x^{3}}\right)$
$\therefore \left(x^{4}-\frac{1}{x^{3}}\right)^{15}$
$={ }^{15} C_{0}\left(x^{4}\right)^{15}\left(-\frac{1}{x^{3}}\right)^{0}$
$+{ }^{15} C_{1}\left(x^{4}\right)^{14}\left(-\frac{1}{x^{3}}\right)+{ }^{15} C_{2}\left(x^{4}\right)^{13}\left(-\frac{1}{x^{3}}\right)^{2}$
$+{ }^{15} C_{3}\left(x^{4}\right)^{12}\left(-\frac{1}{x^{3}}\right)^{3}+{ }^{15} C_{4}\left(x^{4}\right)^{11}\left(-\frac{1}{x^{3}}\right)^{4}+...$
$T_{r+1}={ }^{15} C_{r}\left(x^{4}\right)^{15-r}\left(-\frac{1}{x^{3}}\right)^{r}$
$={ }^{-15} C_{r} x^{60-7 r}$
$\Rightarrow x^{60-7 r}=x^{32}$
$\Rightarrow 60-7 r=32$
$\Rightarrow 7 r=28$
$\Rightarrow r=4$
So, $5th$ term, contains $x^{32}$
$={ }^{15} C_{4}\left(x^{4}\right)^{11}\left(-\frac{1}{x^{3}}\right)^{4}$
$={ }^{15} C_{4} x^{44} x^{-12}={ }^{15} C_{4} x^{32} x^{32}$
$={ }^{15} C_{4}$