College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 9 - Section 9.3 - Geometric Sequences; Geometric Series - 9.3 Assess Your Understanding: 15

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 $\sqrt[3]{2}$.
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