Question

Prove that the coefficient of $x^p$ in the expansion of ${(x^2+\frac{1}{x})}^{2n}$ is $\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$

Answer 1

In the expansion of ${\left(x^{2+}\frac{1}{x}\right)}^{2n}$ we have

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

$=2n{C}_r.{\left(x\right)}^{4n-2r}.\frac{1}{x^r}$

$=2n{C}_r.{\left(x\right)}^{4n-3r}$

Putting $(4n-3r)$=p we get $r=\frac{1}{3}(4n-p)$

Coefficient of $x^p=2n{C}_{\left(\frac{1}{3}\right)(4n-p)}$

$=\frac{2n!}{\left[\frac{1}{3}\right(4n-p)!]\ast [2n-\frac{1}{3}(4n-p\left)\right]!}$

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