Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 14 - Sequences, Series, and the Binomial Theorem - 14.1 Sequences and Series - 14.1 Exercise Set - Page 894: 28


$a_{1}=\frac{(1)^2-1}{(1)^2+1}=\frac{0}{2}=0$ $a_{2}=\frac{(2)^2-1}{(2)^2+1}=\frac{3}{5}$ $a_{3}=\frac{(3)^2-1}{(3)^2+1}=\frac{8}{10}$ $a_{4}=\frac{(4)^2-1}{(4)^2+1}=\frac{15}{17}$ $a_{10}=\frac{(10)^2-1}{(10)^2+1}=\frac{99}{101}$ $a_{15}=\frac{(15)^2-1}{(15)^2+1}=\frac{224}{226}$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=\frac{(1)^2-1}{(1)^2+1}=\frac{0}{2}=0$ $a_{2}=\frac{(2)^2-1}{(2)^2+1}=\frac{3}{5}$ $a_{3}=\frac{(3)^2-1}{(3)^2+1}=\frac{8}{10}$ $a_{4}=\frac{(4)^2-1}{(4)^2+1}=\frac{15}{17}$ $a_{10}=\frac{(10)^2-1}{(10)^2+1}=\frac{99}{101}$ $a_{15}=\frac{(15)^2-1}{(15)^2+1}=\frac{224}{226}$
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.