Answer
The sequence is geometric with a common ratio of $\dfrac{2}{3}$.
The first four terms are:
$u_1=2$
$u_2=\frac{4}{3}$
$u_3=\frac{8}{9}$
$u_4=\frac{16}{27}$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$u_1=\dfrac{2^1}{3^{1-1}}=\dfrac{2}{3^0}=\dfrac{2}{1}=2$
$u_2=\dfrac{2^2}{3^{2-1}}=\dfrac{4}{3^1}=\dfrac{4}{3}$
$u_3=\dfrac{2^3}{3^{3-1}}=\dfrac{8}{3^2}=\dfrac{8}{9}$
$u_4=\dfrac{2^4}{3^{4-1}}=\dfrac{16}{3^3}=\dfrac{16}{27}$
To check if the sequence is geometric, find the ratio of each pair of successive terms:
$\dfrac{a_2}{a_1}=\dfrac{\frac{4}{3}}{2}=\dfrac{4}{3} \cdot \dfrac{1}{2}=\dfrac{2}{3}$
$\dfrac{a_3}{a_2}=\dfrac{\frac{8}{9}}{\frac{4}{3}}=\dfrac{8}{9} \cdot \dfrac{3}{4}=\dfrac{2}{3}$
$\dfrac{a_4}{a_3}=\dfrac{\frac{16}{27}}{\frac{8}{9}}=\dfrac{16}{27} \cdot \dfrac{9}{8}=\dfrac{2}{3}$
The common ratios are the same, so the sequence is geometric with a common ratio of $\frac{2}{3}$.
