Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 43

Answer

First five terms are $0,3,8,15,24$

Work Step by Step

Given, $a_{n} = n^{2} -1$ Substituting $n=1,2,3,4,5$ we get, $a_{1} = 1^{2} -1 = 1 - 1 = 0$ $a_{2} = 2^{2} -1 = 4 - 1 = 3$ $a_{3} = 3^{2} -1 = 9 - 1 = 8$ $a_{4} = 4^{2} -1 = 16 - 1 = 15$ $a_{5} = 5^{2} -1 = 25 - 1 = 24$ First five terms are $0,3,8,15,24$
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