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). 
 A: If you are not sure on how to compute $(0.2121\ldots)_{3}$, we have: $$\begin{align} E = (0.\color{red}{2}\color{green}{1}\color{red}{2}\color{green}{1}\ldots)_3 = \frac{1}{3}\times \color{red}{2} + \frac{1}{3^2}\times \color{green}{1} + \frac{1}{3^3}\times \color{red}{2} + \frac{1}{3^4}\times \color{green}{1} + \ldots \\ = \frac{2}{3} + \frac{1}{9} + \frac{2}{27} + \frac{1}{81} + \ldots \\ = \frac{7}{9}+ \frac{7}{81}+\ldots \end{align}$$

Now note that: $$\begin{align} E = \frac{7}{9}+\frac{7}{81}+\frac{7}{729}+\ldots \\ =\frac{7}{9}+\frac{7}{9}\times \frac{1}{9} + \frac{7}{9}\times [\frac{1}{9}]^2\ldots \\ =\frac79(1+\frac19 + [\frac19]^2+\ldots)\end{align}$$
Now, this is an infinite geometric progression. Can you compute the required sum?
A: In base $3$ the digits $21$ represent the number $2 \times 3 + 1 = 7$. 
In base $3$ moving the "decimal point" right two spaces is dividing by $3 \times 3 = 9$. Adding up your infinite decimal two places at a time tells you the sum is
$$
\frac{7}{9}\left( 1 + \frac{1}{9}   + \frac{1}{81} + \cdots \right) = 
\frac{7}{9} \times \frac{1}{1 - 1/9} .
$$
A: You have
$(0.212121 \ldots)_3 = \frac{7}{9} + \frac{7}{81} + \frac{7}{729} + \dots$
and for $0<a<1$
$1+a+a^2+\ldots = \dfrac{1}{1-a}$
from which 
$\begin {align}
(0.212121 \ldots)_3 &= 7\left(\frac{1}{9} + \frac{1}{81} + \frac{1}{729} + \dots 
 \right)\\
 &= 7\left(\frac{1}{1-\frac{1}{9}} -1 \right)\\
 &= 7\left(\frac{9}{8} -1 \right)\\
 &= \frac{7}{8} \\
\end{align}$

You can also (perhaps this is what you actually did) multiply by $100_3 = 9_{\text{dec}}$ to see that for $x=0.\overline{21}_3$, $9x-x = 21_3 = 7_{\text{dec}}$ and solve from there.
A: \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}
