Chapter 9.1, Problem 90E

a

To determine

To find :The third partial sum of the series ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$ .

Answer to Problem 90E

  $\frac{26}{27}$

Explanation of Solution

Given series: ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$

Calculation:

Given series can be written as

  ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}=2\left(\frac{1}{3}\right)+2{\left(\frac{1}{3}\right)}^2+2{\left(\frac{1}{3}\right)}^3+2{\left(\frac{1}{3}\right)}^4+\mathrm{......}$

We know that nth partial sum of a series is the sum of first n terms of the series.

Therefore, the third partial sum of given series is

  $S_3=2\left(\frac{1}{3}\right)+2{\left(\frac{1}{3}\right)}^2+2{\left(\frac{1}{3}\right)}^3$

which is a geometric series of 3 terms with first term $a=\frac{2}{3}$ and common ratio $r=\frac{1}{3}$ .

The sum of n of a geometric series with first term a and common ratio r is

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

Therefore, the third partial sum of given series is

  $S_3=\frac{\frac{2}{3}\left(1-{\left(\frac{1}{3}\right)}^3\right)}{1-\frac{1}{3}}=\frac{\frac{2}{3}\left(1-\frac{1}{27}\right)}{\frac{2}{3}}=\frac{26}{27}$

b

To determine

To find :The fourth partial sum of the series ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$ .

Answer to Problem 90E

  $\frac{80}{81}$

Explanation of Solution

Given series: ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$

Calculation:

The fourth partial sum of given series is

  $\begin{array}{l}{S}_4=2\left(\frac{1}{3}\right)+2{\left(\frac{1}{3}\right)}^2+2{\left(\frac{1}{3}\right)}^3+2{\left(\frac{1}{3}\right)}^4\\ S_4=\frac{\frac{2}{3}\left(1-{\left(\frac{1}{3}\right)}^4\right)}{1-\frac{1}{3}}=\frac{\frac{2}{3}\left(1-\frac{1}{81}\right)}{\frac{2}{3}}=\frac{80}{81}\end{array}$

c

To determine

To find :The fifth partial sum of the series ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$ .

Answer to Problem 90E

  $\frac{242}{243}$

Explanation of Solution

Given series: ${\displaystyle \sum _{i=1}^{\infty }2{\left(\frac{1}{3}\right)}^i}$

Calculation:

The fifth partial sum of given series is

  $\begin{array}{l}{S}_5=2\left(\frac{1}{3}\right)+2{\left(\frac{1}{3}\right)}^2+2{\left(\frac{1}{3}\right)}^3+2{\left(\frac{1}{3}\right)}^4+2{\left(\frac{1}{3}\right)}^5\\ S_5=\frac{\frac{2}{3}\left(1-{\left(\frac{1}{3}\right)}^5\right)}{1-\frac{1}{3}}=\frac{\frac{2}{3}\left(1-\frac{1}{243}\right)}{\frac{2}{3}}=\frac{242}{243}\end{array}$