Chapter 12.3, Problem 44AYU

To determine

To find: The sum of series.

Answer to Problem 44AYU

The sum of series is $S_n=5\left[1-{\left(\frac{3}{5}\right)}^n\right]$ .

Explanation of Solution

Given information:

The series is $2+\frac{6}{5}+\frac{18}{25}+\frac{2^3}{4}+\mathrm{............}+2{\left(\frac{3}{5}\right)}^{n-1}$ .

Calculation:

For the given series $a_1=2$ , $a_2=\frac{6}{5}$ and $a_3=\frac{18}{25}$ .

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

  $\begin{array}{c}\frac{a_2}{a_1}=\frac{\frac{6}{5}}{2}\\ =\frac{3}{5}\end{array}$

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

  $\begin{array}{c}\frac{a_3}{a_1}=\frac{\frac{18}{25}}{\frac{6}{5}}\\ =\frac{3}{5}\end{array}$

Since the ratio of successive terms is same it is a geometric series with common ratio $r=\frac{3}{5}$ .

The sum of first $n$ terms of a geometric sequence for $r<1$ is calculated by,

  $S_n=a_1\frac{1-r^n}{1-r}$

Substitute the given values.

  $\begin{array}{l}{S}_n=2\left(\frac{1-{\left(\frac{3}{5}\right)}^n}{1-\frac{3}{5}}\right)\\ S_n=2\frac{\left(1-{\left(\frac{3}{5}\right)}^n\right)}{\frac{2}{5}}\\ S_n=5\left[1-{\left(\frac{3}{5}\right)}^n\right]\end{array}$

Therefore, the sum of series is $S_n=5\left[1-{\left(\frac{3}{5}\right)}^n\right]$ .