Question

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

Answer 1

The general term in the expansion of ${(x^2+\frac{1}{x})}^{2n}$ is given by
$T_{r+1}{=}_r^{2n}\times (x^2{)}^{(2n-r)}\times {\left(\frac{1}{x}\right)}^r\Rightarrow T_{r+1}{=}^{2n}{C}_r\times x^{(4n-3r)}$
This term contains $x^p$ only when 4n - 3r = P.
And $4n-3r=p\Rightarrow r=\frac{(4n-p)}{3}$
$\text{Putting}4n-3r=p\text{in (i), we get}\text{Coefficeint of}{x}^p{=}^{2n}{C}_r\text{, where r}=\frac{(4n-p)}{3}=\frac{\left(2n\right)!}{(r!)\times (2n-r)!}=\frac{\left(2n\right)!}{\{\left(\frac{4n-p}{3}\right)!\}\times \left\{[2n-\frac{(4n-p)}{3}]\right\}!}=\frac{\left(2n\right)!}{\{\left(\frac{4n-p}{3}\right)!\}\times \left\{\left(\frac{2n+p}{3}\right)\right\}!}\text{hence the coefficient of}{x}^p\text{in the expansion of}{(x^2+\frac{1}{x})}^{2n}\text{is}\frac{\left(2n\right)!}{\left\{(\frac{4n-p}{3}!)\right\}\times \left\{\left(\frac{2n+p}{3}\right)\right\}!}$