## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

We need to substitute $1, 2, 3,$ and $4$ for $n$ into the given equation to find the first four terms. $d_1=\dfrac{3^{1}}{9}=\dfrac{1}{3}$ $d_2=\dfrac{3^{2}}{9}=1$ $d_3=\dfrac{3^3}{9}=\dfrac{27}{9}=3$ $d_4=\dfrac{3^{4}}{4}=\dfrac{81}{9}=9$ We see that the the next term is equal to $3$ times the current term and this implies that the sequence is geometric with a common ratio of $3$.