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

Question

Dr. Abubacker · Accepted Answer

(b)
Since, $n$ is even, therefore $\left( \frac{n}{2} + 1 ight)$th term is the middle term.
$	herefore T_{\frac{n}{2} + 1} = \ ^{n}C_{\frac{n}{2}}\left( x^{2} ight)^{\frac{n}{2}}\left( \frac{1}{x} ight)^{\frac{n}{2}}$
$= 924x^{6}$ (given)
$\Rightarrow \ x^{\frac{n}{2}} = x^{6}\  \Rightarrow \ n = 12$

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

Explanation

Reviewed by Dr. Abubacker

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