## College Algebra 7th Edition

Published by Brooks Cole

# Chapter 8, Sequences and Series - Section 8.1 - Sequences and Summation Notation - 8.1 Exercises - Page 600: 6

#### Answer

$a_1=0$ $a_2=3$ $a_3=8$ $a_4=15$ $a_{100}=9999$

#### Work Step by Step

We are given: $a_{n}=n^{2}-1$ We evaluate: $a_1=(1)^{2}-1=1-1=0$ $a_2=(2)^{2}-1=4-1=3$ $a_3=(3)^{2}-1=9-1=8$ $a_4=(4)^{2}-1=16-1=15$ $a_{100}=(100)^{2}-1=9999$

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