Question

If $x^p$ occurs in the expansion of ${(x^2+\frac{1}{x})}^{2n}.$ then prove that its coefficient is $\frac{2n!}{\frac{(4n-p)!}{3!}\frac{(2n+p)!}{3!}}.$

Answer 1

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

$T_{r+1}=n_{C_r}{a}^{n-r}{x}^r$

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

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

$={2n}_{C_r}{x}^{4n-2r}.\frac{1}{x^r}$

$={2n}_{C_r}{x}^{4n-3r}$

We need the coefficient of $x^p$

$\therefore$ $4n-3r=p$

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

$\therefore$ Coefficient of $x^p$ in ${(x^2+\frac{1}{x})}^{2n}$ is

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

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

Proved