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

Toolbox:
• $T_{r+1}=nC_r(a)^{n-r}b^r$
In the expansion of $(x^2+\large\frac{1}{x})^{2n}$ we have
$t_{r+1}=2nC_r.(x^2)^{2n-r}.(\large\frac{1}{x})^r$
$\Rightarrow 2nC_r .x^{4n-2r} .\large\frac{1}{x^r}$
$\Rightarrow 2nC_r .x^{4n-3r}$
Putting $(4n-3r)=p$ we get