Q85 / 180 · COMEDK 2024 (Memory Based) · Mathematics
Middle term of the expansion $\left(x^2+\frac{1}{x^2}+2\right)^n$ is
- Correct answerA${ }^{2 n} C_n$
- B${ }^{2 n} C_{n+1}$
- C${ }^{4 n} C_n$
- D${ }^{4 n} C_{n+1}$
Solution
We have, $\left(x^2+\frac{1}{x^2}+2\right)^n=\left(x+\frac{1}{x}\right)^{2 n}$
Coefficient of middle term of $(x+y)^n$, if $n$ is even $={ }^n C_{n / 2}$
$\Rightarrow$ Coefficient of middle term of $\left(x+\frac{1}{x}\right)^{2 n}$
$={ }^{2 n} C_{\frac{2 n}{2}}={ }^{2 n} C_n$
Coefficient of middle term of $(x+y)^n$, if $n$ is even $={ }^n C_{n / 2}$
$\Rightarrow$ Coefficient of middle term of $\left(x+\frac{1}{x}\right)^{2 n}$
$={ }^{2 n} C_{\frac{2 n}{2}}={ }^{2 n} C_n$
Hence, the correct answer is option: 'A'.