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: 26

Answer

$a_{1}=(1)^2-2(1)=1-2=-1$ $a_{2}=(2)^2-2(2)=4-4=0$ $a_{3}=(3)^2-2(3)=9-6=3$ $a_{4}=(4)^2-2(4)=16-8=8$ $a_{10}=(10)^2-2(10)=100-20=80$ $a_{15}=(15)^2-2(15)=225-30=195$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(1)^2-2(1)=1-2=-1$ $a_{2}=(2)^2-2(2)=4-4=0$ $a_{3}=(3)^2-2(3)=9-6=3$ $a_{4}=(4)^2-2(4)=16-8=8$ $a_{10}=(10)^2-2(10)=100-20=80$ $a_{15}=(15)^2-2(15)=225-30=195$
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