Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 9 - Section 9.6 - Conic Sections in Polar Coordinates - Exercise Set - Page 1032: 59

Answer

$-\frac{1}{2}$, $\frac{1}{8}$, $-\frac{1}{28}$, $\frac{1}{80}$

Work Step by Step

Given the formula $a_n=\frac{(-1)^n}{3^n-1}$, we have: (i) $a_1=\frac{(-1)^1}{3^1-1}=-\frac{1}{2}$ (ii) $a_2=\frac{(-1)^2}{3^2-1}=\frac{1}{8}$ (iii) $a_3=\frac{(-1)^3}{3^3-1}=-\frac{1}{28}$ (iv) $a_4=\frac{(-1)^4}{3^4-1}=\frac{1}{80}$
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