Question

Find the sum of the first $20$terms of the geometric sequence$3,6,12,24,48,96,\dots$.

Short answer

The sum of the first $20$terms is$\begin{array}{rcl}& & 3\left(2^{20}-1\right)\end{array}$.

Step-by-step solution

Step 1. Given information

The geometric sequence is$3,6,12,24,48,96,...$.

Step 2. Find the first and the common ratio.

Consider the sequence $3,6,12,24,48,96,...$

Let $a,r$denote the first term and the common ratio respectively.

$\begin{array}{rcl}a& =& 3\\ r& =& \frac{6}{3}\\ & =& 2\end{array}$

Step 3. Find the sum of the first 20 terms of the geometric sequence. 

The sum of $n$terms of the sequence is given by $S_n=\frac{a\left(1-r^n\right)}{1-r}$.

Put $a=3,r=2$.

$\begin{array}{rcl}{S}_{20}& =& \frac{3\left(1-2^{20}\right)}{1-2}\\ & =& 3\left(2^{20}-1\right)\end{array}$

$=3145725$