The middle term in the expansion of
$\left(1 + \dfrac{1}{x^2}\right)\!\left(1 + x^2\right)^n$ is —
🎥 Video solution / Text Solution of this question is given below:
Expand $\left(1 + x^2\right)^n = \sum_{k=0}^{n} {}^{n}C_{k}x^{2k}.$
Multiply by $\left(1 + \dfrac{1}{x^2}\right)$:
$= \sum_{k=0}^{n} {}^{n}C_{k}x^{2k} + \sum_{k=0}^{n} {}^{n}C_{k}x^{2k-2}.$
To find middle term → powers of $x$ that are equal when $2k = 2n - (2k-2)$.
Simplifying gives $k = n.$
So middle term = ${}^{2n}C_{n}.$