Question

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

• A. $10$
• B. $12$
• C. $14$
• D. $noneofthese$

Answer 1

The correct option is B $12$

For ${(x^2+\frac{1}{x})}^n$ have middle term.

n is even

$\Rightarrow n=2a;aEN$

middle term ${}^{2a}{C}_a(x^2{)}^a.\frac{1}{x^a}{=}^{2a}{C}_a.x^a$

$=924x^6$

$\Rightarrow a=6$

Also for $a=6{,}^{2a}{C}_a{=}^{12}{C}_6=924$

So, $n=2a=12$