# How do you find the sum of Sigma (3^n+4^n)/5^n from n is [0,oo)?

Feb 18, 2017

${\sum}_{n = 0}^{\infty} \frac{{3}^{n} + {4}^{n}}{5} ^ n = \frac{15}{2}$

#### Explanation:

${\sum}_{n = 0}^{\infty} \frac{{3}^{n} + {4}^{n}}{5} ^ n = {\sum}_{n = 0}^{\infty} \left({3}^{n} / {5}^{n} + {4}^{n} / {5}^{n}\right)$
$\text{ } = {\sum}_{n = 0}^{\infty} {3}^{n} / {5}^{n} + {\sum}_{n = 0}^{\infty} {4}^{n} / {5}^{n}$
$\text{ } = {\sum}_{n = 0}^{\infty} {\left(\frac{3}{5}\right)}^{n} + {\sum}_{n = 0}^{\infty} {\left(\frac{4}{5}\right)}^{n}$

The 1st sum is a GP with 1st term $a = 1$, and common ratio $r = \frac{3}{5}$
The 2nd sum is a GP with 1st term $a = 1$, and common ratio $r = \frac{4}{5}$

Using the GP formula: ${S}_{n} = \frac{a}{1 - r}$, we get:

${\sum}_{n = 0}^{\infty} \frac{{3}^{n} + {4}^{n}}{5} ^ n = \left(\frac{1}{1 - \frac{3}{5}}\right) + \left(\frac{1}{1 - \frac{4}{5}}\right)$
$\text{ } = \left(\frac{1}{\frac{2}{5}}\right) + \left(\frac{1}{\frac{1}{5}}\right)$
$\text{ } = \frac{5}{2} + 5$
$\text{ } = \frac{15}{2}$