Question

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

• A. $n=11$
• B. $n=13$
• C. $n=12$
• D. $n=14$

Answer 1

Answer: (C)

$n=12$

$T_{r+1}={}^n{C}_r{x}^{2n-3r}$
Therefore power of $x$ in the middle term is $6$.
Hence, $2n-3r=6$
$⟹r=\frac{2n}{3}-2$
The coefficient is $924$.
Therefore, ${}^n{C}_{\frac{2n}{3}-2}=924$
$⟹\frac{n!}{(\frac{n}{3}+2)!(\frac{2n}{3}-2)!}=924$
If we substitute $n=12$ (since it is divisible by 3),
We get ${}^n{C}_{\frac{2n}{3}-2}$
$={}^{12}{C}_6$
$=924$
Hence $n=12$