Answer
The sequence is geometric with a common ratio of $3$.
The first four terms are:
$d_1=\frac{1}{3}$
$d_2=1$
$d_3=3$
$d_4=9$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$d_1=\dfrac{3^1}{9} = \dfrac{3}{9}=\dfrac{1}{3}$
$d_2=\dfrac{3^2}{9} = \dfrac{9}{9}=1$
$d_3=\dfrac{3^3}{9} = \dfrac{27}{9}=3$
$d_4=\dfrac{3^4}{9} = \dfrac{81}{9}=9$
Notice that the next term is equal to $3$ times the current term.
This means that a common ratio of $3$ exists, and the sequence is geometric.