We have a random variable $X$ that is distributed according to the Poisson distribution with $ \lambda$, such that $P[X=k]=e^{-\lambda}\frac{ \lambda^k}{k!}$ for $k=0,1,\dots$.
We want to show that probability that $X$ takes on an even value is $\frac{1+e^{-2\lambda}}{2}$.
I am thinking that the identities $e^{t}=\sum_{k=0}^{\infty}\frac{t^k}{k!}$ and $e^{-t}=\sum_{k=0}^{\infty}\frac{-t^{k}}{k!}$ could be used, namely through a system of equations, but I need help figuring out how to productively do this.