Find the exact sum of
$\displaystyle \sum_{i = 0}^{\inf} \frac{4+6^n}{8^n}$
because its an infinite series i know i have to use the formula
$\displaystyle \frac{a}{1-x}$
but im not sure about the general term...
$\displaystyle \sum\limits_{n = 0}^\infty {\frac{{4 + {6^n}}}
{{{8^n}}}} = 4\sum\limits_{n = 0}^\infty {\frac{1}
{{{8^n}}}} + \sum\limits_{n = 0}^\infty {\frac{{{6^n}}}
{{{8^n}}}} = 4\sum\limits_{n = 0}^\infty {\frac{1}
{{{8^n}}}} + \sum\limits_{n = 0}^\infty {{{\left( {\frac{3}
{4}} \right)}^n}} .$
$\displaystyle 4\sum\limits_{n = 0}^\infty {\frac{1}{{{8^n}}}} = 4 \cdot \frac{1}{{1 - {1 \mathord{\left/{\vphantom {1 8}} \right.\kern-\nulldelimiterspace} 8}}} = 4 \cdot \frac{8}{7} = \frac{{32}}{7}.$
$\displaystyle \sum\limits_{n = 0}^\infty {{{\left( {\frac{3}{4}} \right)}^n}} = \frac{1}{{1 - {3 \mathord{\left/{\vphantom {3 4}} \right.
\kern-\nulldelimiterspace} 4}}} = 4.$
$\displaystyle \sum\limits_{n = 0}^\infty {\frac{{4 + {6^n}}}{{{8^n}}}} = \frac{{32}}{7} + 4 = \frac{{60}}{7}.$