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If $n$ is even and the middle term in the expansion of $\left(x^2+\frac{1}{x}\right)^n$ is $924 x^6$, then $n$ is equal to
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The correct answer is:
$12$
For $\left(x^2+\frac{1}{x}\right)^n$ have middle term $n$ is even
$\Rightarrow n=2 a ; a \in N$
Middle term ${ }^{2 a} C_a\left(x^2\right)^a \cdot \frac{1}{x^a}={ }^{2 a} C_a x^a=924 x^6$
$\Rightarrow \quad a=6$
Also, for $a=6,{ }^{2 a} C_a={ }^{12} C_6=924$
So, $n=2 a=12$
$\Rightarrow n=2 a ; a \in N$
Middle term ${ }^{2 a} C_a\left(x^2\right)^a \cdot \frac{1}{x^a}={ }^{2 a} C_a x^a=924 x^6$
$\Rightarrow \quad a=6$
Also, for $a=6,{ }^{2 a} C_a={ }^{12} C_6=924$
So, $n=2 a=12$
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