Answer
$\dfrac{1}{6}(3^n-1)$
Work Step by Step
We are given the sum:
$\dfrac{3}{9}+\dfrac{3^2}{9}+\dfrac{3^3}{9}+....+\dfrac{3^n}{9}$
Consider the geometric sequence:
$a_1=\dfrac{3}{9}$
$r=3$
The given sum represents the sum $S_n$ of the first $n$ terms. Determine this sum:
$S_n=a_1\dfrac{1-r^n}{1-r}=\dfrac{3}{9}\cdot\dfrac{1-3^n}{1-3}$
$=\dfrac{1}{6}(3^n-1)$