Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 11 - Further Topics in Algebra - 11.1 Sequences and Series - 11.1 Exercises: 19

Answer

$a_{1}=1;a_{2}=\dfrac {7}{6};a_{3}=1;a_{4}=\dfrac {5}{6};a_{5}=\dfrac {19}{27}$

Work Step by Step

$a_{1}=\dfrac {4\times 1-1}{1^{2}+2}=1;$ $a_{2}=\dfrac {4\times 2-1}{2^{2}+2}=\dfrac {7}{6};$ $a_{3}=\dfrac {4\times 3-1}{3^{2}+2}=1$ $a_{4}=\dfrac {4\times 4-1}{4^{2}+2}=\dfrac {15}{18}=\dfrac {5}{6};$ $a_{5}=\dfrac {4\times 5-1}{5^{2}+2}=\dfrac {19}{27}$
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