Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.3 Arithmetic Sequences - 9.3 Exercises: 13

Answer

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

Work Step by Step

$a_n=\frac{(-1)^n}{n^2}$ $a_1=\frac{(-1)^1}{1^2}=-1$ $a_2=\frac{(-1)^2}{2^2}=\frac{1}{4}$ $a_3=\frac{(-1)^3}{3^2}=\frac{-1}{9}$ $a_4=\frac{(-1)^4}{4^2}=\frac{1}{16}$ $a_5=\frac{(-1)^5}{5^2}=\frac{-1}{25}$
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.