4.4.1 problem 62

$\text{E}\left(2^X\right)={\sum }_{k=0}^{\infty }{2}^{k}P(X=k)=e^{-\lambda }{\sum }_{k=0}^{\infty }\frac{{\left(2\lambda \right)}^k}{k!}=e^{-\lambda }{e}^{2\lambda }=e^{\lambda }$.