Chapter 12, Problem 17RE

To determine

To find:Whether the series is arithmetic, geometric or neither, if the series is arithmetic find the common difference and sum of first $n$ terms and if the series is geometric find the common ratio and the sum of first $n$ terms.

Answer to Problem 17RE

The sequence is geometric, common ratio is $8$ and the sum of the $n$ terms is $\frac{8}{7}\left(1-8^n\right)$ .

Explanation of Solution

Given: $\left\{s_n\right\}=\left\{2^{3n}\right\}$

The terms of the series are,

  $\begin{array}{l}{s}_1=2^{3\left(1\right)}=8\\ s_2=2^{3\left(2\right)}=64\\ s_3=2^{3\left(3\right)}=512\\ s_4=2^{3\left(4\right)}=4096\end{array}$

If the series is arithmetic then it must the follow the relation,.

  $\begin{array}{c}{s}_2-s_1=s_3-s_2\\ 64-8\ne 512-64\end{array}$

The sequence is not arithmetic.

Foe the sequence to be geometric it must follow,

  $\begin{array}{c}\frac{s_2}{s_1}=\frac{s_3}{s_2}\\ \frac{64}{8}=\frac{512}{64}\\ 8=8\end{array}$

So, the sequence is geometric.

The common ratio is $8$ .

The sum is calculated as,’

  $\begin{array}{c}T=8\left(\frac{1-8^n}{1-8}\right)\\ =\frac{8}{7}\left(1-8^n\right)\end{array}$

Therefore, the sequence is geometric, common ratio is $8$ and the sum of the $n$ terms is $\frac{8}{7}\left(1-8^n\right)$