# Thread: sum of an infinite series

1. ## sum of an infinite series

Find the exact sum of
$\sum_{i = 0}^{\inf} \frac{4+6^n}{8^n}$

because its an infinite series i know i have to use the formula
$\frac{a}{1-x}$
but im not sure about the general term...

2. Originally Posted by acosta0809
Find the exact sum of
$\sum_{i = 0}^{\inf} \frac{4+6^n}{8^n}$

because its an infinite series i know i have to use the formula
$\frac{a}{1-x}$
but im not sure about the general term...

$\sum\limits_{n = 0}^\infty {\frac{{4 + {6^n}}}
{{{8^n}}}} = 4\sum\limits_{n = 0}^\infty {\frac{1}
{{{8^n}}}} + \sum\limits_{n = 0}^\infty {\frac{{{6^n}}}
{{{8^n}}}} = 4\sum\limits_{n = 0}^\infty {\frac{1}
{{{8^n}}}} + \sum\limits_{n = 0}^\infty {{{\left( {\frac{3}
{4}} \right)}^n}} .$

$4\sum\limits_{n = 0}^\infty {\frac{1}{{{8^n}}}} = 4 \cdot \frac{1}{{1 - {1 \mathord{\left/{\vphantom {1 8}} \right.\kern-\nulldelimiterspace} 8}}} = 4 \cdot \frac{8}{7} = \frac{{32}}{7}.$

$\sum\limits_{n = 0}^\infty {{{\left( {\frac{3}{4}} \right)}^n}} = \frac{1}{{1 - {3 \mathord{\left/{\vphantom {3 4}} \right.
\kern-\nulldelimiterspace} 4}}} = 4.$

$\sum\limits_{n = 0}^\infty {\frac{{4 + {6^n}}}{{{8^n}}}} = \frac{{32}}{7} + 4 = \frac{{60}}{7}.$