# Which equation could be used to calculate the sum of the geometric series?

$\text{Series} = \dfrac{1}{3}+ \dfrac{2}{9}+ \dfrac{4}{27}+ \dfrac{8}{21}+ \dfrac{16}{243}$

This problem aims to familiarize us with the arrangement 0f object in series and sequences. The concepts required to solve this problem include geometric series and geometric sequences. The main difference between a series and a sequence is that there exists an arithmetic operation in sequence whereas a series is just a series of objects separated by a comma.

There are several examples of sequences but here we are going to use the geometric sequence, which is a sequence where, every ascending term is acquired by using arithmetic operations of multiplication or division, on a real number with the previous number. The sequence is written in the form:

$a, ar, ar^2, ……., ar^{n-1}, …..$

The method used here is $\dfrac{\text{Succesive term}}{\text{preceding term}}$.

Whereas to find the sum of the first $n$ terms, we use the formula:

$S_n = \dfrac{a(1-r^n)}{(1-r)} \space if\space r<1$

$S_n = \dfrac{a(r^n-1)}{(r-1)} \space if\space r>1$

Here, $a = \text{first term}$, $r = \text{common ratio}$, and $n = \text{term position}$.

First, we have to determine the common ratio of the series, as it will indicate which formula is to be applied. So the common ratio of a series is found by dividing any term by its previous term:

$r = \dfrac{\text{Succesive term}}{\text{preceding term}}$

$r = \dfrac{2}{9} \div \dfrac{1}{3}$

$r = \dfrac{2}{3}\space r < 1$

Since $r$ is less than $1$, we will be using:

$S_n = \dfrac{a_1(1-r^n)}{(1-r)} \space if\space r<1$

We have $a_1 = \dfrac{1}{3}$, $n = 5$ terms, and $r = \dfrac{2}{3}$, substituing them in the above equation gives us:

$S_5 = \dfrac{\dfrac{1}{3}(1-(\dfrac{2}{3})^5)}{(1-\dfrac{2}{3})}$

$S_5 = \dfrac{\dfrac{1}{3}(1-(\dfrac{32}{243}))}{(\dfrac{3-2}{3})}$

$S_5 = \dfrac{\dfrac{1}{3}(\dfrac{243-32}{243})}{(\dfrac{1}{3})}$

$S_5 = \dfrac{\dfrac{1}{3}\times \dfrac{211}{243}}{\dfrac{1}{3}}$

$S_5 = \dfrac{\cancel{\dfrac{1}{3}}\times \dfrac{211}{243}}{\cancel{\dfrac{1}{3}}}$

$S_5 = \dfrac{211}{243}$

## Numerical Result

The equation $S_n = \dfrac{a_1(1-r^n)}{(1-r)} \space if\space r<1$ is used to calculate the sum, and the sum is $S_5 = \dfrac{211}{243}$.

## Example

Find the common ratio and the first four terms of the geometric sequence

$\{\dfrac{2^{n-3}}{4}\}$.

The simplest part of solving this problem is calculating the first four terms of the sequence. This is can be done by plugging in the numbers $1, 2, 3,$ and $4$ into the formula given in the problem.

The first term can be found by plugging $1$ into the equation:

$a_1 = \dfrac{2^{1-3}}{4} = \dfrac{2^{-2}}{4} = \dfrac{1}{2^2\times 4}$

$a_1 = \dfrac{1}{4\times 4} = \dfrac{1}{16}$

The second term can be found by plugging $2$ into the equation:

$a_2 = \dfrac{2^{2-3}}{4} = \dfrac{2^{-1}}{4} = \dfrac{1}{2^1\times 4}$

$a_2 = \dfrac{1}{2\times 4} = \dfrac{1}{8}$

The third term can be found by plugging $3$:

$a_3=\dfrac{2^{3-3}}{4} = \dfrac{2^0}{4} =\dfrac{1}{4}$

The fourth and the last term can be found by plugging $4$:

$a_4=\dfrac{2^{4-3}}{4} = \dfrac{2^{1}}{4} = \dfrac{2^1}{4}$

$a_4=\dfrac{2}{4} = \dfrac{1}{2}$

The series is: $\dfrac{1}{16}, \dfrac{1}{8}, \dfrac{1}{4}, \dfrac{1}{2}, …$

The common ratio can be found by:

$r=\dfrac{\text{Succesive term}}{\text{preceding term}}$

$r=\dfrac{1}{16} \div \dfrac{1}{8}$

$r=\dfrac{1}{2}$