The middle term in the expansion of $\left(1 + \dfrac{1}{x^2}\right)\!\left(1 + x^2\right)^n$ is —

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 = ${}^{2n}C_{n}.$