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Middle Terms

If n is even ...

Question

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

A

10

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B

12

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C

14

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D

6

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Solution

The correct option is B $12$
$∵ n$ is even
$∴$ The middle term $= {(\frac{n}{2} + 1)}^{th}$ term

$\Rightarrow T_{(n / 2 + 1)} =^{n} C_{n / 2} (x^{2})^{n / 2} {(\frac{1}{x})}^{n / 2}$

$= 924 x^{6}$ (given)

$\Rightarrow x^{n / 2} = x^{6} \Rightarrow n = 12$

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