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: 32

Answer

$a_{1}=(-1)^1(1)^2=-1$ $a_{2}=(-1)^2(2)^2=4$ $a_{3}=(-1)^3(3)^2=-9$ $a_{4}=(-1)^4(4)^2=16$ $a_{10}=(-1)^{10}(10)^2=100$ $a_{15}=(-1)^{15}(15)^2=-225$

Work Step by Step

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