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: 20

Answer

$a_{1}=0;$ $a_{2}=\dfrac {3}{5};$ $a_{3}=\dfrac {4}{5};$ $a_{4}=\dfrac {15}{17};$ $a_{5}=\dfrac {12}{13}$

Work Step by Step

$a_{1}=\dfrac {1^{2}-1}{1^{2}+1}=0;$ $a_{2}=\dfrac {2^{2}-1}{2^{2}+1}=\dfrac {3}{5};$ $a_{3}=\dfrac {3^{2}-1}{3^{2}+1}=\dfrac {4}{5};$ $a_{4}=\dfrac {4^{2}-1}{4^{2}+1}=\dfrac {15}{17};$ $a_{5}=\dfrac {5^{2}-1}{5^{2}+1}=\dfrac {24}{26}=\dfrac {12}{13}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.