Chapter 9, Problem 54RE

To determine

To find : the sum of the finite geometric sequence

Answer to Problem 54RE

  ${\displaystyle \sum _{i=1}^6{\left(\frac{1}{3}\right)}^{i-1}}=\frac{364}{243}$

Explanation of Solution

Given information: ${\displaystyle \sum _{i=1}^6{\left(\frac{1}{3}\right)}^{i-1}}$

Calculation:

Given sum can be written as

  $\begin{array}{l}{\displaystyle \sum _{i=1}^6{\left(\frac{1}{3}\right)}^{i-1}}={\left(\frac{1}{3}\right)}^{1-1}+{\left(\frac{1}{3}\right)}^{2-1}+{\left(\frac{1}{3}\right)}^{3-1}+\mathrm{....}+{\left(\frac{1}{3}\right)}^{6-1}\\ {\displaystyle \sum _{i=1}^6{\left(\frac{1}{3}\right)}^{i-1}}=1+\frac{1}{3}+\frac{1}{3^2}+\mathrm{......}+\frac{1}{3^5}\end{array}$

which is finite geometric series with 6 terms . The first term of the series is $a=1$ and common ratio is $r=\frac{1}{3}$ .

We know that the sum to n terms of the geometric series is

  $S_n=\frac{a\left(1-r^n\right)}{1-r}$

Therefore, we have

  $S_6=\frac{1\left[1-{\left(\frac{1}{3}\right)}^6\right]}{1-\frac{1}{3}}=\frac{\left[1-\frac{1}{729}\right]}{\frac{2}{3}}=\frac{364}{243}$