How do I find the function represented by the following series?

Question

$$\sum_{k=0}^{\infty}\frac{2(-1)^k x^{k+2}}{k!}$$

I'm supposed to find the function represented by the power series but I'm really stumped... (x is raised to the k+2 power)

Answer 1

If you’re allowed to use the fact that $e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$ for all x, then you can write

\begin{align*} \sum_{k=0}^{\infty}\frac{2(-1)^k x^k+2}{k!} &= \sum_{k=0}^{\infty}2\cdot\frac{(-x)^k+1}{k!}\\ &= 2\sum_{k=0}^{\infty}\left(\frac{(-x)^k}{k!}+\frac{1}{k!}\right)\\ &= 2\sum_{k=0}^{\infty}\frac{(-x)^k}{k!}+2\sum_{k=0}^{\infty}\frac{1}{k!}\\ &= 2e^{-x}+2e \end{align*}

Answer 2

Factor out $2x^2$ and rewrite the rest as a series of the form $$\sum_{k=0}^\infty \frac{z^k}{k!} = e^z.$$