Factor out $e^{-\lambda}$ and notice that you have even series from a Taylor expansion, which is equivalent to $0.5*(e^x + e^{-x})$.
Edit To make this more detailed, notice that in the Taylor expansion
$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$
you would like to keep even terms and remove the odd terms. Note that
$e^{-x} = \sum_{k=0}^\infty \frac{(-1)^k x^k}{k!} = \sum_{k \text{even}}\frac{x^k}{k!} - \sum_{k \text{ odd}}\frac{x^k}{k!},$
since $(-1)^k$ is $1$ for even $k$ and $-1$ for odd $k$. Now you can add the two series, and the odd terms will cancel, exactly as you need. The problem is, you will have each even twice, not once. To deal with that, divide by $2$, getting
$\frac{e^x + e^{-x}}{2} = \sum_{k \text{even}}\frac{x^k}{k!} = \sum_{i = 0}^\infty \frac{x^(2i)}{(2i)!},$
as desired, since the left-hand side is exactly $\cosh x$.