University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 3

Answer

$a_1=1$ $a_2=\frac{-1}{3}$ $a_3=\frac{1}{5}$ $a_4=\frac{-1}{7}$

Work Step by Step

We know that $a_n=\frac{(-1)^{n+1}}{2n-1}$. Hence, here: $a_1=\frac{(-1)^{1+1}}{2\cdot1-1}=\frac{(-1)^{2}}{2-1}=\frac{1}{1}=1$ $a_2=\frac{(-1)^{2+1}}{2\cdot2-1}=\frac{(-1)^{3}}{4-1}=\frac{-1}{3}$ $a_3=\frac{(-1)^{3+1}}{2\cdot3-1}=\frac{(-1)^{4}}{6-1}=\frac{1}{5}$ $a_4=\frac{(-1)^{4+1}}{2\cdot4-1}=\frac{(-1)^{5}}{8-1}=\frac{-1}{7}$
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