First slide

Binomial theorem for positive integral Index

Question

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

Moderate

A

10
B

12
C

14
D

None of these

Solution

Since, n is even, therefore $(\frac{n}{2} + 1)$ th term is the middle term.
$\begin{aligned} ∴ T_{\frac{n}{2} + 1} =^{n} C_{n / 2} {(x^{2})}^{n / 2} {(\frac{1}{x})}^{n / 2} = 924 x^{6} (g i v e n) \\ \Rightarrow x^{n / 2} = x^{6} \Rightarrow n = 12 \end{aligned}$

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