Chapter 11.6, Problem 3ST

(a)

To determine

To calculate:

The sum value of the following geometric series:

  $n=6,r=2,t_1=5$

Answer to Problem 3ST

The sum of the geometric series is $315$ .

Explanation of Solution

Given information:

  $n=6,r=2,t_1=5$

Formula used:

The sum of the geometric series is given by the following expression:

  $S_n=\frac{t_1\left(1-r^n\right)}{1-r}$

Calculation:

The sum of the geometric series is calculated as:

  $S_n=\frac{t_1\left(1-r^n\right)}{1-r}$

Put the given data as:

  $\begin{array}{l}{S}_n=\frac{5\left(1-2^6\right)}{1-2}\\ S_n=\frac{5\left(1-64\right)}{-1}\\ S_n=\frac{-315}{-1}\\ S_n=315\end{array}$

(b)

To determine

To calculate:

The sum value of the following geometric series:

  ${{\displaystyle \sum _{k=1}^4\left(\frac{2}{3}\right)}}^k$

Answer to Problem 3ST

The sum of the geometric series is $1.6049$ .

Explanation of Solution

Given information:

  ${{\displaystyle \sum _{k=1}^4\left(\frac{2}{3}\right)}}^k$

Calculation:

The summation notation of the geometric series is given as:

  $\begin{array}{l}{{\displaystyle \sum _{k=1}^4\left(\frac{2}{3}\right)}}^k={\left(\frac{2}{3}\right)}^1+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^3+{\left(\frac{2}{3}\right)}^4\\ {{\displaystyle \sum _{k=1}^4\left(\frac{2}{3}\right)}}^k=\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{64}{81}\\ {{\displaystyle \sum _{k=1}^4\left(\frac{2}{3}\right)}}^k=1.6049\end{array}$