Answer
The sequence is geometric with a common ratio of $\dfrac{2}{3}$.
Work Step by Step
We need to substitute $1, 2, 3,$ and $4$ for $n$ into the given equation to find the first four terms.
$u_1=\dfrac{2^{1}}{3^{(1-1)}}=\dfrac{2}{1}=2 $
$u_2=\dfrac{2^{2}}{3^{(2-1)}}=\dfrac{4}{3^1}=\dfrac{4}{3}$
$u_3=\dfrac{2^{3}}{3^{(3-1)}}=\dfrac{8}{3^2}=\dfrac{8}{9}$
$u_4=\dfrac{2^{4}}{3^{(4-1)}}=\dfrac{16}{3^3}=\dfrac{16}{27}$
Dividing the first two terms gives:
$\frac{4}{3}\div 2=\frac{2}{3}$
Dividing the other terms gives the same result.
We can see that the common ratios are the same; thus the sequence is geometric with a common ratio of $\dfrac{2}{3}$.
