Chapter 9.3, Problem 60E

To determine

To find: the sum of the finite geometric sequence.

Answer to Problem 60E

The sum of the finite geometric sequence is 12.4999.

Explanation of Solution

Given information:

The given sequence,

  ${\displaystyle \sum _{n=0}^{40}5{\left(\frac{3}{5}\right)}^n}$

Concept used:

The sum of the finite geometric sequence

  $a_1,a_{1}r,a_1{r}^2,a_1{r}^3,\mathrm{.................}{a}_1{r}^{n-1}$ with common ratio $r\ne 1$ is given by

  $\begin{array}{c}{S}_n={\displaystyle \sum _{i=1}^n{a}_1{r}^{i-1}}\\ =a\left(\frac{1-r^n}{1-r}\right)\end{array}$

Calculation:

Consider the finite geometric sequence

  ${\displaystyle \sum _{n=0}^{40}5{\left(\frac{3}{5}\right)}^n}$

By writing few terms,

  ${\displaystyle \sum _{n=0}^{40}{\left(\frac{3}{5}\right)}^n}=5{\left(\frac{3}{5}\right)}^0+5{\left(\frac{3}{5}\right)}^1+5{\left(\frac{3}{5}\right)}^2+\mathrm{.......}+5{\left(\frac{3}{5}\right)}^{40}$

Now because $a_1=5,r=\frac{3}{5}$ and $n=41$ , apply the formula for the sum of a finite geometric sequence to obtain

  $\begin{array}{c}{S}_n=a_1\left(\frac{1-r^n}{1-r}\right)\\ {\displaystyle \sum _{n=0}^{40}5{\left(\frac{3}{5}\right)}^n}=5\left(\frac{1-{\left(\frac{3}{5}\right)}^{41}}{1-\left(\frac{3}{5}\right)}\right)\\ \approx 12.4999\end{array}$