Let’s also notice that the initial value of the index is \(n = 0\) and so we can put this into the form,

\[\sum\limits_{n = 0}^\infty {a\,{r^n}} \]

At that point we’ll be able to determine if it converges or diverges and the value of the series if it does happen to converge.

In this case it’s pretty simple to put the series into the form above so here is that work.

\[\sum\limits_{n = 0}^\infty {{3^{2 + n}}\,{2^{1 - 3n}}} = \sum\limits_{n = 0}^\infty {{3^2}{3^n}\,{2^1}{2^{ - 3n}}} = \sum\limits_{n = 0}^\infty {\left( 9 \right)\left( 2 \right)\frac{{{3^n}}}{{{2^{3n}}}}} = \sum\limits_{n = 0}^\infty {18\frac{{{3^n}}}{{{8^n}}}} = \sum\limits_{n = 0}^\infty {18{{\left( {\frac{3}{8}} \right)}^n}} \]

Make sure you properly deal with any negative exponents that might happen to be in the terms!

Also recall that all the exponents must be simply \(n\) and can’t be 3\(n\) or anything else. So, for this problem, we’ll need to use basic exponent rules to write \({2^{3n}} = {\left( {{2^3}} \right)^n} = {8^n}\).

Show Step 3

With the series in “proper” form we can see that \(a = 18\) and \(r = \frac{3}{8}\). Therefore, because we can clearly see that \(\left| r \right| = \frac{3}{8} < 1\), the series will converge and its value is,

\[\sum\limits_{n = 0}^\infty {{3^{2 + n}}\,{2^{1 - 3n}}} = \sum\limits_{n = 0}^\infty {18{{\left( {\frac{3}{8}} \right)}^n}} = \frac{{18}}{{1 - \frac{3}{8}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{144}}{5}}}\]