The sum of the infinite series $\dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dfrac{2^8}{8!} + \dots$ is equal to —

We know that $e^x = 1 + \dfrac{x}{1!} + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dots$ Let $S = \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots$ Write using even terms of $e^x$: $e^2 = 1 + 2 + \dfrac{2^2}{2!} + \dfrac{2^3}{3!} + \dfrac{2^4}{4!} + \dots$ and $e^{-2} = 1 - 2 + \dfrac{2^2}{2!} - \dfrac{2^3}{3!} + \dfrac{2^4}{4!} - \dots$ Adding both, $e^2 + e^{-2} = 2\left(1 + \dfrac{2^2}{2!} + \dfrac{2^4}{4!} + \dfrac{2^6}{6!} + \dots\right)$ So, $S = \dfrac{1}{2}\left(e^2 + e^{-2} - 2\right) = \dfrac{1}{2}\left(\dfrac{e^4 + 1 - 2e^2}{e^2}\right) = \dfrac{(e^2 - 1)^2}{2e^2}.$