### What is the sum of the first $5$ terms of the geometric series $1, \dfrac{ 2 }{ 3 }, \dfrac{ 4 }{ 9 } \space ...?$

$\dfrac{ 211 }{ 81 }$

Step by Step Explanation:
1. The sum of first $n$ terms of a $G.P.$ is given by,
$S_n = \dfrac{ a(1 - r^n) }{ (1 - r) }$
Here, the first term, $a = 1$ and
the common ratio, $r = \dfrac{ a_{k+1} }{ a_k } \text{ where } k ≥ 1$
$\implies r = \dfrac{ \dfrac{ 2 }{ 3 } } { 1 } = \dfrac{ 2 }{ 3 }$
2. The sum of first $n$ terms of this $G.P.$ is given by,
\begin{align} S_n &= \dfrac { (1)\left(1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right) } { 1 - \dfrac{ 2 }{ 3 } } \\ &= \dfrac { \left(1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right) } { \dfrac{ 1 }{ 3 } } \\ &= 3 \left[ 1 - \left( \dfrac{ 2 }{ 3 } \right)^n \right] \\ \end{align} Now, the sum of the first $5$ terms of the $G.P$ is given by, \begin{align} S_5 &= 3 \left[ 1 - \left( \dfrac{ 2 }{ 3 } \right)^5 \right] \\ &= 3 \left[ 1 - \dfrac{ 32 }{ 243 } \right] \\ &= 3 \times \dfrac{ 211 }{ 243 } \\ &= \dfrac{ 211 }{ 81 } \end{align}
3. Hence, the sum of the first $5$ terms of the $G.P.$ is $\dfrac{ 211 }{ 81 }$.