Chapter 9, Problem 52RE

To determine

Sum of a finite geometric sequence

Answer to Problem 52RE

The sum of the finite geometric series is 121

Explanation of Solution

Given:

Given function:

  ${\displaystyle \sum _{i=1}^5{3}^{i-1}}$

Consider the following

  ${\displaystyle \sum _{i=1}^5{3}^{i-1}}$

The sum of the finite geometric sequence $S_n={\displaystyle \sum _{i=1}^n{a}_1{r}^{i-1}}$ when $r\ne 1$ is,

  $S_n=a_1\left(\frac{1-r^n}{1-r}\right)$

Here, $a_1$ the first is term and r is common ratio

Write down few terms of the sequence,

  ${\displaystyle \sum _{i=1}^5{3}^{i-1}}=3^0+3^1+3^2+3^3+\mathrm{...}+3^4$

Now

  $a_1=1$

And

  $a_2=3$

So common ratio is,

  $\begin{array}{l}r=\frac{a_2}{a_1}\hfill \\ =\frac{3}{1}\hfill \\ =3\hfill \end{array}$

The number of terms in the sequence are n=5

Then,

  $\begin{array}{l}{\displaystyle \sum _{i=1}^5{3}^{i-1}}=1\left(\frac{1-3^5}{1-3}\right)\\ =\frac{242}{2}\\ =121\end{array}$

Hence,

  ${\displaystyle \sum _{i=1}^5{3}^{i-1}}=121$