Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises: 15

Answer

$a_{1}=0,a_{2}=\dfrac {1}{9};a_{3}=\dfrac {2}{27};a_{4}=\dfrac {1}{27};a_{5}=\dfrac {4}{243}$

Work Step by Step

$a_{1}=\left( \dfrac {1}{3}\right) ^{1}\left( 1-1\right) =0;$ $a_{2}=\left( \dfrac {1}{3}\right) ^{2}\left( 2-1\right) =\dfrac {1}{9};$ $a_{3}=\left( \dfrac {1}{3}\right) ^{2}\left( 3-1\right) =\dfrac {2}{27};$ $a_{4}=\left( \dfrac {1}{3}\right) ^{4}\left( 4-1\right) =\dfrac {1}{27}$ $a_{5}=\left( \dfrac {1}{3}\right) ^{5}\left( 5-1\right) =\dfrac {4}{243}$
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