Answer
${{s}_{n}}=\frac{1-{{\left( -x \right)}^{n}}}{1+x}$.
Work Step by Step
The sum of the $n\text{th}$ term of the geometric series is given by
${{s}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}$, $r\ne 1$ …… (1)
Where ${{a}_{1}}$ is the first term, $n$ is the number of terms, and $r$ is the factor between the terms (common ratio).
The common ratio of geometric series is $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$
Substitute $n=1$ in the equation $r=\frac{{{a}_{n+1}}}{{{a}_{n}}}$
$\begin{align}
& r=\frac{{{a}_{n+1}}}{{{a}_{n}}} \\
& =\frac{{{a}_{2}}}{{{a}_{1}}}
\end{align}$
$\begin{align}
& r=\frac{-x}{1} \\
& =-x
\end{align}$
To find the value of ${{s}_{n}}$, substitute the value of ${{a}_{1}}=1$ and $r=-x$ in equation (1)
$\begin{align}
& {{s}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r} \\
& =\frac{1\left\{ 1-{{\left( -x \right)}^{n}} \right\}}{1-\left( -x \right)} \\
& =\frac{1-{{\left( -x \right)}^{n}}}{1+x}
\end{align}$
