Answer
The sequence is geometric with a common ratio of $\sqrt[3]{2}$.
The first four terms are:
$e_1=\sqrt[3]{2}$
$e_2=\sqrt[3]{4}$
$e_3=2$
$e_4=2\sqrt[3]{2}$
Work Step by Step
Substitute $1, 2, 3,$ and $4$ for $n$ into the given formula:
$e_1=2^{1/3}=2^{1/3}=\sqrt[3]{2}$
$e_2=2^{2/3}=\sqrt[3]{2^2} = \sqrt[3]{4}$
$e_3=2^{3/3}=2^1=2$
$e_4=2^{4/3}=\sqrt[3]{2^4} = \sqrt[3]{2^3(2)}=2\sqrt[3]{2}$
To check if the sequence is geometric, find the ratio of each successive pairs:
$\dfrac{a_2}{a_1}=\dfrac{\sqrt[3]{4}}{\sqrt[3]{2}} =\sqrt[3]{\dfrac{4}{2}}=\sqrt[3]{2}$
$\dfrac{a_3}{a_2}=\dfrac{2}{\sqrt[3]{4}}=\dfrac{2 \cdot \sqrt[3]{2}}{\sqrt[3]{4} \cdot \sqrt[3]{2}}=\dfrac{2\sqrt[3]{2}}{\sqrt[3]{8}}=\dfrac{2\sqrt[3]{2}}{2}=\sqrt[3]{2}$
$\dfrac{a_4}{a_3}=\dfrac{2\sqrt[3]{2}}{2}=\sqrt[3]{2}$
The common ratios are the same, so the sequence is geometric with a common ratio of $\sqrt[3]{2}$.