Answer
$\frac{547}{18}$
Work Step by Step
The series is $\frac{1}{18}-\frac{1}{6}+\frac{1}{2}-\cdots $.
Thus, ${{a}_{1}}=\frac{1}{18}$, $n=7$ and
$\begin{align}
& r=\frac{-\frac{1}{6}}{\frac{1}{18}} \\
& =-\frac{1}{6}\cdot \frac{18}{1} \\
& =-3
\end{align}$
Substituting ${{a}_{1}}=\frac{1}{18}$, $n=7$ and $r=-3$ in ${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}$, for any $r\ne 1$:
$\begin{align}
& {{S}_{9}}=\frac{\frac{1}{18}\left( 1-{{\left( -3 \right)}^{7}} \right)}{1-\left( -3 \right)} \\
& =\frac{\frac{1}{18}\left( 1+2187 \right)}{4} \\
& =\frac{\frac{1}{18}\cdot \left( 2188 \right)}{4} \\
& =\frac{547}{18}
\end{align}$
Thus, the sum of the first nine terms, ${{S}_{7}}$ for $\frac{1}{18}-\frac{1}{6}+\frac{1}{2}-\cdots $ is $\frac{547}{18}$.
