Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 735: 7

Answer

$1,-\frac{1}{4},\frac{1}{9},-\frac{1}{16},\frac{1}{25}$

Work Step by Step

$a_{n}=\frac{(-1)^{n-1}}{n^{2}}$ Substitute $1,2,3,4,5$ to $n$: $a_{1}=\frac{(-1)^{1-1}}{1^{2}}=1$ $a_{2}=\frac{(-1)^{2-1}}{2^{2}}=-\frac{1}{4}$ $a_{3}=\frac{(-1)^{3-1}}{3^{2}}=\frac{1}{9}$ $a_{4}=\frac{(-1)^{4-1}}{4^{2}}=-\frac{1}{16}$ $a_{5}=\frac{(-1)^{5-1}}{5^{2}}=\frac{1}{25}$
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