University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.2 - Infinite Series - Exercises - Page 498: 89


a) $2 +2(\dfrac{3}{5})+2(\dfrac{3}{5})^2+....$ b) $\dfrac{13}{2}-\dfrac{13}{2}(\dfrac{3}{10})+\dfrac{13}{2}(\dfrac{3}{10})^2+....$

Work Step by Step

a) Here, $\dfrac{2}{1-r}=5$ This implies $1-r=\dfrac{2}{5} \implies r=\dfrac{3}{5}$ Thus, we will have the series: $2 +2(\dfrac{3}{5})+2(\dfrac{3}{5})^2+....$ b) Here, $\dfrac{\dfrac{13}{2}}{1-r}=5$ This implies $1-r=\dfrac{13}{10} \implies r=\dfrac{-3}{10}$ Thus, we will have series: $\dfrac{13}{2}-\dfrac{13}{2}(\dfrac{3}{10})+\dfrac{13}{2}(\dfrac{3}{10})^2+....$
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