Question

In Problems 13–16, find each sum.

$\sum _{k=1}^7{\left(\frac{1}{3}\right)}^k$

Short answer

The sum is:$\frac{1}{2}\left(1-3^{-7}\right)$

Step-by-step solution

Step 1. Given information. 

The summation notation of the sequence is:

$\sum _{k=1}^7{\left(\frac{1}{3}\right)}^k$

Step 2. Expand the summation by substituting values of k.

$\sum _{k=1}^7{\left(\frac{1}{3}\right)}^k={\left(\frac{1}{3}\right)}^1+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+{\left(\frac{1}{3}\right)}^4+.......{\left(\frac{1}{3}\right)}^7$

Step 3. Check type of sequence.

Here we can see that ratio between successive term is constant.

So given series is a geometric series.

First term is$a=\frac{1}{3}$

common ratio is $$\begin{gathered} r={\left(\frac{1}{3}\right)}^2÷\left(\frac{1}{3}\right) \\ =\frac{1}{3} \end{gathered}$$

Number of terms is$n=7$

Step 4. Find sum of geometric series.

The sum of all 7 terms:

$$\begin{gathered} \frac{1}{3}+{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^3+.....+{\left(\frac{1}{3}\right)}^7=\frac{\frac{1}{3}\left\{1-{\left(\frac{1}{3}\right)}^7\right\}}{1-{\displaystyle \frac{1}{3}}} \\ =\frac{{\displaystyle \frac{1}{3}(1-3^{-7})}}{{\displaystyle \frac{2}{3}}} \\ =\frac{1}{2}(1-3^{-7)} \end{gathered}$$