The sum of n terms of a geometric series is found using the following formula:
\(S _{n} = \frac{a(1-r^{n})}{1-r}\)
where a is the first term and r is the common ratio.
In your question, a = 6 and r = 1/3, so we have:
\(S_{5} =\frac{6(1-(\frac{1}{3})^5)}{1-\frac{1}{3}}\)
which simplifies to:\(S_{5}=\frac{6(1-\frac{1}{243})}{\frac{2}{3}}\)
