Find sum of infinite series $\sum_{k=0}^\infty \frac{2^{k+3}}{e^{k-3}}$

Question

Find the sum of infinite series

$$\sum_{k=0}^\infty \frac{2^{k+3}}{e^{k-3}}$$

I know that the first term is $8e^3$ and the second one $16e^2$ but then I get stuck.

Answer 1

HINT: This is a geometric series because $\dfrac{a_{k+1}}{a_k} = \dfrac{\frac{2^{k+4}}{e^{k-2}}}{\frac{2^{k+3}}{e^{k-3}}} = \dfrac2e$ is a constant. Now, you can use the closed form of geometric series.

Answer 2

$$\frac{2^{k+2}}{e^{k-3}} = \frac{2^k\cdot 2^2}{e^k\cdot e^{-3}} = \frac{2^k}{e^k}\cdot 4e^3$$

Can you use this to rewrite your sum as a geometric sum?