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  4. If the middle term in the expansion of (x2+(1/x))n is 924 x6 then value of n is

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

Binomial Theorem

Solution:

$n$ is even here
$ \therefore $ middle term $={ }^n C_{n / 2}\left(x^2\right)^{\frac{n}{2}} \cdot\left(\frac{1}{x}\right)^{\frac{n}{2}}=924 x^6$
$ \Rightarrow{ }^n C_{n / 2} \cdot x^{n / 2}=924 x^6 $
$ \therefore \frac{n}{2}=6 \Rightarrow n=12$