Step 1:
First, we write out each term of the series:
\begin{align*}
\sum_{j=2}^{7} \frac{1}{3}(4)^{j-1} &= \frac{1}{3}(4)^{2-1} + \frac{1}{3}(4)^{3-1} + \frac{1}{3}(4)^{4-1} + \frac{1}{3}(4)^{5-1} + \frac{1}{3}(4)^{6-1} + \frac{1}{3}(4)^{7-1} \\
&= \frac{1}{3}(4) +
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