rf n is even, then the middle term in the expansion of ${\left({\mathrm{x}}^2+\frac{1}{\mathrm{x}}\right)}^{\mathrm{n}}$is 924 x⁶, then n is equal to

Solution:

Since, n is even, therefore $\left(\frac{\mathrm{n}}{2}+1\right)$th term is the middle term.

$\begin{array}{r}\therefore T_{\frac{n}{2}+1}{=}^n{C}_{n/2}{\left(x^2\right)}^{n/2}{\left(\frac{1}{x}\right)}^{n/2}=924x^6 \left(given\right)\\ \Rightarrow x^{n/2}=x^6\Rightarrow n=12\end{array}$