College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises: 9

Answer

$a_{1}=-1$ $a_{2}=\frac{1}{4}$ $a_{3}=-\frac{1}{9}$ $a_{4}=\frac{1}{16}$ $a_{100}=\frac{1}{10000}$

Work Step by Step

We are given: $a_{n}= \frac{(-1)^{n}}{n^{2}}$ We evaluate: $a_{1}= \frac{(-1)^{1}}{1^{2}}=\frac{-1}{1}=-1$ $a_{2}= \frac{(-1)^{2}}{2^{2}}=\frac{1}{4}$ $a_{3}= \frac{(-1)^{3}}{3^{2}}=-\frac{1}{9}$ $a_{4}= \frac{(-1)^{4}}{4^{2}}=\frac{1}{16}$ $a_{100}= \frac{(-1)^{100}}{100^{2}}=\frac{1}{10000}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.