Answer
$\Sigma_{n=0}^{\infty}\dfrac{(1+2^{n})x^{n}}{n!}$
$R=\infty$
Work Step by Step
$e^{x}+e^{2x}=\Sigma_{n=0}^{\infty}\frac{x^{n}}{n!}+\Sigma_{n=0}^{\infty}\dfrac{(2x)^{n}}{(n)!}$
$=\Sigma_{n=0}^{\infty}\dfrac{(1+2^{n})x^{n}}{n!}$
$R=\infty$