Question

If the middle term in the expansion of ${\left(x^2+\frac{1}{x}\right)}^n$ is $924x^6$ then value of $n$ is

• A. 10
• B. 12
• C. 14
• D. 16

Answer 1

Answer: (B)

$n$ is even here
$\therefore$ middle term $={}^n{C}_{n/2}{\left(x^2\right)}^{\frac{n}{2}}\cdot {\left(\frac{1}{x}\right)}^{\frac{n}{2}}=924x^6$
$\Rightarrow {}^n{C}_{n/2}\cdot x^{n/2}=924x^6$
$\therefore \frac{n}{2}=6\Rightarrow n=12$