Question

If $x^p$ occurs in the expansion of ${(x^2+\frac{1}{x})}^{2n}$, prove that its coefficient is

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

Answer 1

The general term of the expansion $(a+b{)}^n$ is given by

$T_{r+1}{=}^n{C}_r{a}^{n-r}{b}^r$

Given: ${(x^2+\frac{1}{x})}^{2n}$

$T_{r+1}{=}^{2n}{C}_r(x^2{)}^{2n-r}{\left(\frac{1}{x}\right)}^r$

$\Rightarrow T_{r+1}{=}^{2n}{C}_r\left(x{)}^{4n-2r-r}{=}^{2n}{C}_r\right(x{)}^{4n-3r}$

The coefficient of $x^p$ occurs when

$4n-3r=p$

$\Rightarrow r=\frac{4n-p}{3}$

So, the coefficient of $x^p$

=${}^{2n}{C}_{\left(\frac{4n-p}{3}\right)}$

=$\frac{2n!}{(2n-\frac{4n-p}{3})!\left(\frac{4n-p}{3}\right)!}$

$[{\because }^n{C}_r=\frac{n!}{r!(n-r)!}]$

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

Hence, proved