Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 11 - 11.1 - Sequences and Series - 11.1 Exercises - Page 777: 18

Answer

$a_1=3$ $a_2=\frac{12}{11}$ $a_3=\frac{9}{13}$ $a_4=\frac{24}{47}$ $a_5=\frac{15}{37}$

Work Step by Step

$a_n=\frac{6n}{3n^2-1}$ $a_1=\frac{6(1)}{3(1)^2-1}=\frac{6}{3(1)-1}=\frac{6}{2}=3$ $a_2=\frac{6(2)}{3(2)^2-1}=\frac{12}{3(4)-1}=\frac{12}{11}$ $a_3=\frac{6(3)}{3(3)^2-1}=\frac{18}{3(9)-1}=\frac{18}{26}=\frac{9}{13}$ $a_4=\frac{6(4)}{3(4)^2-1}=\frac{24}{3(16)-1}=\frac{24}{47}$ $a_5=\frac{6(5)}{3(5)^2-1}=\frac{30}{3(25)-1}=\frac{30}{74}=\frac{15}{37}$
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