Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 11 - Section 11.1 - Sequences and Summation Notation - Exercise Set: 4

Answer

$a_1 = \frac{1}{3} \\a_2 = \frac{1}{9} \\a_3 = \frac{1}{27} \\a_4 = \frac{1}{81}$

Work Step by Step

To find the first four terms of the given sequence, substitute 1, 2, 3, and 4 to $n$, respectively, then simplify. $a_1 = (\frac{1}{3})^1 = \frac{1}{3} \\a_2 = (\frac{1}{3})^2 = (\frac{1}{3})(\frac{1}{3}) = \frac{1}{9} \\a_3 = (\frac{1}{3})^3 = (\frac{1}{3})(\frac{1}{3})(\frac{1}{3}) = \frac{1}{27} \\a_4 = (\frac{1}{3})^4 = (\frac{1}{3})(\frac{1}{3})(\frac{1}{3})(\frac{1}{3}) = \frac{1}{81}$
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