Question

The middle term in the expansion of ${(x^2+\frac{1}{x})}^n$ is $924x^6$. If $n$ is even, then $n$ is equal to

• A. $10$
• B. $12$
• C. $24$
• D. $6$

Answer 1

The correct option is B $12$

Since $n$ is even,
$\therefore$ The middle term $={(\frac{n}{2}+1)}^{\text{th}}$ term
$T_{(n/2+1)}={}^n{C}_{n/2}(x^2{)}^{n/2}{\left(\frac{1}{x}\right)}^{n/2}={}^n{C}_{n/2}{x}^{n/2}$
$\Rightarrow T_{(n/2+1)}=924x^6$ (given)
$\Rightarrow x^{n/2}=x^6$
$\Rightarrow n=12$