using a base 3 decimal to express as a base 10 fraction using geometric series 
Express $0.\overline{21}_3$ as a base 10 fraction in reduced form.

So I was able to solve it by setting $x=\overline{.21}$, but the solution also briefly mentioned another way using the geometric series:

A quick way to get the answer by using the geometric series is: $(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots = \frac{7}{8}.$

However, I'm having a hard time understanding how to actually use the geometric series (the above answer is not clear to me). 

\begin{align}
   (0.212121 \ldots)_3
   &= \dfrac{21_3}{100_3} 
    + \dfrac{21_3}{(100_3)^2} 
    + \dfrac{21_3}{(100_3)^3} + \cdots\\
   &= \frac 79 + \frac{7}{9^2} + \frac{7}{9^3} + \cdots \\
   &= \frac 79\left( \dfrac{1}{1-\frac 19} \right) \\  
   &= \frac 79 \cdot \frac 98 \\
   &= \frac 78.
\end{align}
You could also use the old shift and subtract trick...
\begin{array}{rcl}
   100_3x &= &21.212121 \ldots_3 \\
   x &= &\phantom{0}0.212121 \ldots_3 \\
\hline
   22_3x &= &21_3 \\
   8x &= &7 \\
   x &= \dfrac 78
\end{array}
