College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 7 - Conic Sections - Exercise Set 7.3 - Page 701: 92


See below.

Work Step by Step

$n=1:\frac{(-1)^1}{3^1-1}=\frac{-1}{3-1}=\frac{-1}{2}$ $n=2:\frac{(-1)^2}{3^2-1}=\frac{1}{9-1}=\frac{1}{8}$ $n=3:\frac{(-1)^3}{3^3-1}=\frac{-1}{27-1}=\frac{-1}{26}$ $n=4:\frac{(-1)^4}{3^4-1}=\frac{1}{81-1}=\frac{1}{80}$
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