## Thomas' Calculus 13th Edition

$4$
Formula to calculate the sum of a geometric series is: $S=\dfrac{a}{1-r}$; Since, we have two series $\sum_{n =1}^{ \infty}\dfrac{2^n+3^n}{4^n}=\sum_{n =1}^{ \infty} (\dfrac{1}{2})^n+\sum_{n =0}^{ \infty} (\dfrac{3}{4})^n$ Here, $a=\dfrac{1}{2},\dfrac{3}{4}$ and common ratios are: $r =\dfrac{1}{2},\dfrac{3}{4}$ Now, Thus, $S=s_1+s_2=\dfrac{1/2}{1-\dfrac{1}{2}}+\dfrac{3/4}{1-\dfrac{3}{4}}$ or, $=1+3$ or, $=4$