Chapter 8.3, Problem 52E

To determine

To find: the sequence of the first five partial sums $S_1,S_2,S_3,S_4$ and $S_5$ for the given geometric sequence.

Answer to Problem 52E

The sequence of first five partial sums is $8,20,38,65,\frac{211}{2}$ respectively.

Explanation of Solution

Given information:

A geometric sequence is given as

  $8,12,18,27,\frac{81}{2},\mathrm{...}$

Formula used:

The sum $S_n$ or ${\displaystyle \sum _{i=1}^n{a}_r{r}^{r-1}}$ of a finite geometric sequence is $a_1\left(\frac{1-r^n}{1-r}\right)$ , where $a_1$ is the first term, $r$ is the common ration, and $n$ is the number of terms in the sequence.

Calculation:

Consider the given geometric sequence.

  $8,12,18,27,\frac{81}{2},\mathrm{...}$

Write out a few terms of the finite sequence as shown:

  $8,12,18,27,\frac{81}{2},\mathrm{...}=8\cdot {\left(\frac{3}{2}\right)}^0+8\cdot {\left(\frac{3}{2}\right)}^1+8\cdot {\left(\frac{3}{2}\right)}^2+\mathrm{...}$

Substitute 8 for $a_1$ , $\left(\frac{3}{2}\right)$ for $r$ , and 1 for $n$ in $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_1=8\left(\frac{1-{\left(\frac{3}{2}\right)}^1}{1-\left(\frac{3}{2}\right)}\right)\\ =8\end{array}$

Substitute 8 for $a_1$ , $\left(\frac{3}{2}\right)$ for $r$ , and 2 for $n$ in $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_2=8\left(\frac{1-{\left(\frac{3}{2}\right)}^2}{1-\left(\frac{3}{2}\right)}\right)\\ =8\cdot \frac{\frac{5}{4}}{\frac{1}{2}}\\ =20\end{array}$

Substitute 8 for $a_1$ , $\left(\frac{3}{2}\right)$ for $r$ , and 3 for $n$ in $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_3=8\left(\frac{1-{\left(\frac{3}{2}\right)}^3}{1-\left(\frac{3}{2}\right)}\right)\\ =8\cdot \frac{\frac{19}{8}}{\frac{1}{2}}\\ =38\end{array}$

Substitute 8 for $a_1$ , $\left(\frac{3}{2}\right)$ for $r$ , and 4 for $n$ in $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_4=8\left(\frac{1-{\left(\frac{3}{2}\right)}^4}{1-\left(\frac{3}{2}\right)}\right)\\ =8\cdot \frac{\frac{65}{16}}{\frac{1}{2}}\\ =65\end{array}$

Substitute 8 for $a_1$ , $\left(\frac{3}{2}\right)$ for $r$ , and 5 for $n$ in $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$ .

  $\begin{array}{c}{S}_3=8\left(\frac{1-{\left(\frac{3}{2}\right)}^5}{1-\left(\frac{3}{2}\right)}\right)\\ =8\cdot \frac{\frac{211}{32}}{\frac{1}{2}}\\ =\frac{211}{2}\end{array}$

Therefore, the sequence of first five partial sums is $8,20,38,65,\frac{211}{2}$ respectively.