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What is the sum of the first $n$ terms of the geometric series $1, \dfrac{ 3 }{ 5 }, \dfrac{ 9 }{ 25 } \space ...?$

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Answer:

$\dfrac{ 5 }{ 2 } \left[ 1 - \left( \dfrac{ 3 }{ 5 } \right)^n \right]$

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, $r = \dfrac{ a_{k+1} }{ a_k } \text{ where } k ≥ 1$
   $\implies r = \dfrac{ \dfrac{ 3 }{ 5 } } { 1 } = \dfrac{ 3 }{ 5 }$
2. The sum of first $n$ terms of this $G.P.$ is given by,
   $$\begin{align} S_n &= \dfrac { (1)\left(1 - \left( \dfrac{ 3 }{ 5 } \right)^n \right) } { 1 - \dfrac{ 3 }{ 5 } } \\ &= \dfrac { \left(1 - \left( \dfrac{ 3 }{ 5 } \right)^n \right) } { \dfrac{ 2 }{ 5 } } \\ &= \dfrac{ 5 }{ 2 } \left[ 1 - \left( \dfrac{ 3 }{ 5 } \right)^n \right] \\ \end{align}$$
3. Hence, the sum of the first $n$ terms of the $G.P.$ is $\dfrac{ 5 }{ 2 } \left[ 1 - \left( \dfrac{ 3 }{ 5 } \right)^n \right]$.

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