I have the following sum:
$$ S = \sum^{\infty}_{n = 1} \frac{1 + 2^n}{3^n} $$
So I tried the following:
$$ S = \sum^{\infty}_{n = 1} \frac{1}{3^n} + \sum^{\infty}_{n = 1} \frac{2^n}{3^n} = \sum^{\infty}_{n = 1} (\frac{1}{3})^n + \sum^{\infty}_{n = 1} (\frac{2}{3})^n$$
And here, applying the formula for each summatory with $a = 1$ and $r_1 = \frac{1}{3} $ and $r_2 = \frac{2}{3}$ gives me $ \frac{3}{2} $ and $3$ respectively. However, $\frac{9}{2}$ is not the correct answer. I would like to know which is the wrong step here, for me it seems everything looks fine.
The series are $$ \sum_{n=1}^\infty\frac1{3^n}=\frac{1/3}{1-1/3}=\frac12, $$ and $$ \sum_{n=1}^\infty\bigg(\frac23\bigg)^n=\frac{2/3}{1-2/3}=2. $$ Most likely you calculated as if the sums started from $n=0$.