## College Algebra (10th Edition)

The sequence is geometric with a common ratio of $\dfrac{2}{3}$. The first four terms are: $u_1=2$ $u_2=\frac{4}{3}$ $u_3=\frac{8}{9}$ $u_4=\frac{16}{27}$
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 $\frac{2}{3}$.