# Prove the following

Question:

If $\mathrm{x}^{p}$ occurs in the expansion of $x^{2}+\frac{1}{x}$ Prove that its coefficient is

$\frac{\mid 2 n}{\frac{4 n-p}{3} \mid \frac{2 n+p}{3}}$

Solution:

Given expression is $\left(x^{2}+\frac{1}{x}\right)^{2 n}$

Using the standard formula above expression can be written as

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

If $x^{\rho}$ occurs in the expansion,

Let $4 n-3 r=p$

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

Coefficient of $x^{p}={ }^{2 n} C_{r}=\frac{(2 n) !}{r !(2 n-r) !}=\frac{(2 n) !}{\left(\frac{4 n-p}{3}\right) !\left(2 n-\frac{4 n-p}{3}\right) !}$

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