Question

Find the sum of the infinite geometric series.

$64+16+4+1+\frac{1}{4}+\frac{1}{16}+...$

Short answer

Sum is$\frac{256}{3}$.

Step-by-step solution

Step 1. Given information

Infinite geometric series is$64+16+4+1+\frac{1}{4}+\frac{1}{16}+...$

Step 2. Find the sum of the infinite geometric series. 

Consider the infinite geometric series:

$64+16+4+1+\frac{1}{4}+\frac{1}{16}+...$

Let $a,r,S_n$denote the first term, ratio of the consecutive terms and the sum.

$\begin{array}{rcl}a& =& 64\\ r& =& \frac{16}{64}\\ & =& \frac{1}{4}\\ S_n& =& \frac{a}{1-r}\\ & =& \frac{64}{1-{\displaystyle \frac{1}{4}}}\\ & =& \frac{64\times 4}{3}\\ & =& \frac{256}{3}\end{array}$