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

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 $n = \text{term position}$.

## Expert Answer

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:

\[r=\dfrac{\text{Succesive term}}{\text{preceding term}} \]

\[r=\dfrac{1}{16} \div \dfrac{1}{8} \]

\[r=\dfrac{1}{2}\]