How to convert the following sum to a geometric series? 
Find $$
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
$$

There are many ways to find that the limit is divergent, but the question explicitly states the sum must be interpreted as a geometric series.  I know how to render the exponential terms to $(n-1)$; it's just the $(6  - \ldots)$ is bugging me -- I can't for the life of me find a way find to isolate the term so I get $a \cdot r^{n-1}$.

Expanding of Ross Millikan's comment, you could try
\begin{align}
\sum_{n = 1}^{\infty} \frac{6 - 2^{2n - 1}}{3^n}
& = \sum_{n = 1}^{\infty} \frac{6}{3^n} - \sum_{n = 1}^{\infty} \frac{2^{2n - 1}}{3^n}
= 2 \cdot \sum_{n = 1}^{\infty} \left( \frac{1}{3} \right)^{n - 1} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n \\
& = 2 \cdot \sum_{n = 0}^{\infty} \left( \frac{1}{3} \right)^{n} - \frac{1}{2} \sum_{n = 1}^{\infty} \left( \frac{4}{3} \right)^n
= 2 \cdot \frac{3}{2} - \frac{1}{2} \left( \sum_{n = 0}^{\infty} \left(\frac{4}{3} \right)^n - 1\right) \\
& = 3 + \frac{1}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n
= \frac{7}{2} - \frac{1}{2} \sum_{n = 0}^{\infty} \left( \frac{4}{3} \right)^n
= - \infty.
\end{align}
