Question

Sum of infinity the following series:
$3+5.\frac{1}{4}+7.\frac{1}{4^2}+.....$

Answer 1

Given,

$a=3$

$d=5-3=2$

$r=\frac{\frac{1}{4}}{1}=\frac{1}{4}$

Formula,

$S_{\infty }=\frac{a}{1-r}+\frac{dr}{(1-r{)}^2}$

$=\frac{3}{1-\frac{1}{4}}+\frac{2\times \frac{1}{4}}{{(1-\frac{1}{4})}^2}$

$=4+\frac{8}{9}$

$S_{\infty }=\frac{44}{9}$