Answer
$2 \cosh x =\Sigma_{n=0}^{\infty} \dfrac{x^n}{n!}+\Sigma_{n=0}^{\infty} \dfrac{(-x)^n}{n!}$
Work Step by Step
We know that the Maclaurin series for $e^x=\Sigma_{n=0}^{\infty} \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+.....$
Here, we have:
$2 \cosh x=\Sigma_{n=0}^{\infty} \dfrac{x^n}{n!}+\Sigma_{n=0}^{\infty} \dfrac{(-x)^n}{n!}=\Sigma_{n=0}^{\infty} \dfrac{2x^{2n}}{(2n)!}\\=2+x^2+\dfrac{x^4}{12}+......$
Thus, our required series is:
$2 \cosh x =\Sigma_{n=0}^{\infty} \dfrac{x^n}{n!}+\Sigma_{n=0}^{\infty} \dfrac{(-x)^n}{n!}$
