Question

Find the sum.

$\underset{i=1}{\overset{15}{\sum 6}}{\left(2\right)}^i$

Short answer

Sum is equal to$\begin{array}{rcl}& & 393204\end{array}$

Step-by-step solution

Step 1. Find the first term and the common ratio.

Consider the sequence $\underset{i=1}{\overset{15}{\sum 6}}{\left(2\right)}^i$

Let role="math" localid="1645297852087" $a_1,r$denote the first term and the common ratio respectively.

$\begin{array}{rcl}{a}_1& =& 6{\left(2\right)}^1\\ & =& 12\\ a_2& =& 6{\left(2\right)}^2\\ & =& 24\\ & & \end{array}$So,

$\begin{array}{rcl}r& =& \frac{24}{12}\\ & =& 2\end{array}$

Step 2. Find the sum.

Consider the sequence $\sum _{i=1}^{15}6{\left(2\right)}^i$.

Sum of the first $n$terms is given by $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$

So,

$\begin{array}{rcl}\sum _{i=1}^{15}6{\left(2\right)}^i& =& \frac{12\left(1-2^{15}\right)}{1-2}\\ & =& -12\left(1-32768\right)\\ & =& 393204\end{array}$