Intermediate Algebra (6th Edition)

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

Chapter 11 - Sections 11.1-11.3 - Integrated Review - Sequences and Series - Page 653: 4

Answer

$\left\{ -4,-1,4,11,20 \right\} $

Work Step by Step

Given that $ a_n=n^2-5 $, then the first 5 terms are \begin{array}{l} a_1=(1)^2-5\\ a_1=1-5\\ a_1=-4 ,\\\\ a_2=(2)^2-5\\ a_2=4-5\\ a_2=-1 ,\\\\ a_3=(3)^2-5\\ a_3=9-5\\ a_3=4 ,\\\\ a_4=(4)^2-5\\ a_4=16-5\\ a_4=11 ,\\\\ a_5=(5)^2-5\\ a_5=25-5\\ a_5=20 .\end{array} Hence, the first 5 terms are $ \left\{ -4,-1,4,11,20 \right\} $.
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