Answer
The sequence is geometric with a common ratio of $\sqrt[3]{2}$.
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.
$e_1=2^{1/3}=2^{1/3}=\sqrt[3]{2} \\ e_2=2^{2/3}=\sqrt[3]{2^2} = \sqrt[3]{4} \\ e_3=2^{3/3}=2 \\ e_4=2^{4/3}=\sqrt[3]{2^4} = \sqrt[3]{2^3(2)}=2\sqrt[3]{2}$
Our aim is to check if the sequence is geometric and then compute the ratio of each successive pairs.
$\dfrac{a_2}{a_1}=\dfrac{\sqrt[3]{4}}{\sqrt[3]{2}} =\sqrt[3]{2}$
$\dfrac{a_3}{a_2}=\dfrac{2}{\sqrt[3]{4}}=\dfrac{2 \cdot \sqrt[3]{2}}{\sqrt[3]{4} \cdot \sqrt[3]{2}}=\dfrac{2\sqrt[3]{2}}{2}=\sqrt[3]{2}$
$\dfrac{a_4}{a_3}=\dfrac{2\sqrt[3]{2}}{2}=\sqrt[3]{2}$
We can see that the common ratios are the same; thus the sequence is geometric with a common ratio of $\sqrt[3]{2}$.
