Question

The coefficient of $x^P$ in the expansion of ${(x^2+\frac{1}{x})}^{2n}$ is

• A. $\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n-p}{3}\right)!}$.
• B. $\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$.
• C. $\frac{\left(2n\right)!}{\left(\frac{4n+p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$.
• D. $\frac{\left(n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$.

Answer 1

The correct option is B $\frac{\left(2n\right)!}{\left(\frac{4n-p}{3}\right)!\left(\frac{2n+p}{3}\right)!}$.

We are given a binomial expression, ${(x^2+\frac{1}{x})}^{2x}$. Now compare it with

$(a+b{)}^n$. we get $a=x^2$ &$c=\frac{1}{x}$

Here the power is in $2n$

For the coefficient of $x^P$ we have to expand the binomial as below.

So, ${(x^2+\frac{1}{x})}^{2n}=\sum _{r=0}^{2n}{}^{2n}{C}_r.{\alpha }^{2n-r}.b^r$

$=\left(\frac{2n}{0}\right)(x^2{)}^{2n}{\left(\frac{1}{x}\right)}^o+\left(\frac{2n}{1}\right)(x^2{)}^{2n+1}\left(\frac{1}{x}\right)+....+\left(\frac{2n}{2n-1}\right)(x^2{)}^1{\left(\frac{1}{x}\right)}^{2n-1}+\left(\frac{2n}{2n}\right)(x^2{)}^o{\left(\frac{1}{x}\right)}^{2n}$

Now the coefficient of $(r+1{)}^{th}$ term will be given as

$\left(\frac{2n}{r}\right)(x^2{)}^{2n-r}{\left(\frac{1}{x}\right)}^r$

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

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

Let compare the power $x$ to the $P$

i.e. power of $x$

So, $4n-3r=P$

So, $r=\frac{4n-P}{3}$

So, the coefficient of $x^P$ is given by

$=\left(\frac{2n}{r}\right)$

$=\frac{2n!}{r!(2n-r)!}$

But here $r=\frac{4n-P}{3}$ So,

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

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

The coefficient of $x^P$ is

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