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: 28

Answer

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

Work Step by Step

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