# Question #7cea5

Apr 11, 2017

$\frac{9}{2}$

#### Explanation:

${\sum}_{n = 0}^{\infty} \frac{1 + {2}^{n}}{3} ^ n$
$= {\sum}_{n = 0}^{\infty} \frac{1}{3} ^ n + {2}^{n} / {3}^{n}$
$= {\sum}_{n = 0}^{\infty} \left({\left(\frac{1}{3}\right)}^{n} + {\left(\frac{2}{3}\right)}^{n}\right)$
$= {\sum}_{n = 0}^{\infty} {\left(\frac{1}{3}\right)}^{n} + {\sum}_{n = 0}^{\infty} {\left(\frac{2}{3}\right)}^{n}$

Then, use the sum of the geometric series formula: ${\sum}_{i = 0}^{\infty} a {r}^{i} = \frac{a}{1 - r} , | r | < 1$ for both sums.

$= \frac{1}{1 - \frac{1}{3}} + \frac{1}{1 - \frac{2}{3}}$
$= \frac{3}{2} + 3$
$= \frac{9}{2}$