Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 3

Answer

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

Work Step by Step

Substitute $n=1,2,3,4$ in the formula for $a_n$: $a_1$ = $\frac{(-1)^2}{2-1}$ = $1$ $a_2$ =$\frac{(-1)^3}{4-1}$ = $-\frac {1}{3}$ $a_3$ =$\frac{(-1)^4}{6-1}$ = $\frac{1}{5}$ $a_4$ = $\frac{(-1)^5}{8-1}$ = $-\frac{1}{7}$
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