Chapter 12.3, Problem 41AYU

To determine

To find: The sum of series.

Answer to Problem 41AYU

The sum of series is $S_n=2\left[1-{\left(\frac{2}{3}\right)}^n\right]$ .

Explanation of Solution

Given information:

The series is ${\displaystyle \sum _{k=1}^n{\left(\frac{2}{3}\right)}^k}$

Calculation:

Calculate first term.

  $\begin{array}{l}{\displaystyle \sum _{k=1}^1{\left(\frac{2}{3}\right)}^1}=a_1\\ =\frac{2}{3}\end{array}$

Calculate second term.

  $\begin{array}{c}{\displaystyle \sum _{k=1}^n{\left(\frac{2}{3}\right)}^k}=a_2\\ =\left({\frac{2}{3}}^2\right)\end{array}$

Calculate third term.

  $\begin{array}{c}{\displaystyle \sum _{k=1}^n{\left(\frac{2}{3}\right)}^k}=a_3\\ ={\left(\frac{2}{3}\right)}^3\end{array}$

Calculate $\frac{a_2}{a_1}$ .

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{{\left(\frac{2}{3}\right)}^2}{\frac{2}{3}}\\ =\frac{2}{3}\end{array}$

Calculate $\frac{a_3}{a_1}$ .

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{{\left(\frac{2}{3}\right)}^3}{{\left(\frac{2}{3}\right)}^2}\\ =\frac{2}{3}\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=\frac{2}{3}$ .

The sum of first $n$ terms of a geometric sequence is calculated by,

  $S_n=a_1\frac{1-r^n}{1-r}$

Substitute the given values.

  $\begin{array}{l}{S}_n=\frac{2}{3}\left(\frac{1-{\left(\frac{2}{3}\right)}^n}{1-\frac{2}{3}}\right)\\ S_n=\frac{2}{3}\frac{\left(1-{\left(\frac{2}{3}\right)}^n\right)}{\frac{1}{3}}\\ S_n=2\left[1-{\left(\frac{2}{3}\right)}^n\right]\end{array}$

Therefore, the sum of series is $S_n=2\left[1-{\left(\frac{2}{3}\right)}^n\right]$ .