Question

If $x^{4r}$ occurs in the expansion of ${(x+\frac{1}{x^2})}^{4n}$ , then its coefficient is

• A. $\frac{4n!}{\left(\frac{4}{3}\right(n-r\left)\right)!\left(\frac{4}{3}\right(2n+r\left)\right)!}$
• B. $\frac{4n!}{\left(\frac{3}{4}\right(n-r\left)\right)!\left(\frac{3}{4}\right(2n+r\left)\right)!}$
• C. $\frac{4n!}{\left(\frac{4}{3}\right(2n-r\left)\right)!\left(\frac{4}{3}\right(2n+r\left)\right)!}$
• D. None of these

Answer 1

Answer: (A)

$\frac{4n!}{\left(\frac{4}{3}\right(n-r\left)\right)!\left(\frac{4}{3}\right(2n+r\left)\right)!}$

The $(k+1{)}^{th}$ term of the given expression is
$T_{k+1}={}^{4n}{\text{C}}_k{x}^{(4n-k)}(\frac{1}{x^2}{)}^k$

We know the coefficient of $x$ is $4r$, so, $4n-k-2k=4r$
$k=\frac{4(n-r)}{3}$

The $(k+1{)}^{th}$ term will be,
${}^{4n}{\text{C}}_{\frac{4(n-r)}{3}}\left(x{)}^{(4n-\frac{4(n-r)}{3})}\right(\frac{1}{x^2}{)}^{\frac{4(n-r)}{3}}$

The coeffecient will be ${}^{4n}{\text{C}}_{\frac{4(n-r)}{3}}$
That is,
= $\frac{4n!}{(4n-\frac{4n}{3}+\frac{4r}{3})!(\frac{4(n-r)}{3})!}$

= $\frac{4n!}{(\frac{4(2n+r)}{3})!(\frac{4(n-r)}{3})!}$