Q83 / 180 · KCET 2023 · Mathematics

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

  • A
    $14$
  • Correct answer
    B
    $12$
  • C
    $8$
  • D
    $10$

Solution

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, $n=2 a=12$

Hence, the correct answer is option: 'B'.

Relevant Questions