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: 34

Answer

$a_{1}=(-1)^{1+1}(3(1)-5)=(-1)^2(-2)=-2$ $a_{2}=(-1)^{2+1}(3(2)-5)=(-1)^3(1)=-1$ $a_{3}=(-1)^{3+1}(3(3)-5)=(-1)^4(4)=4$ $a_{4}=(-1)^{4+1}(3(4)-5)=(-1)^5(7)=-7$ $a_{10}=(-1)^{10+1}(3(10)-5)=(-1)^{11}(25)=-25$ $a_{15}=(-1)^{15+1}(3(15)-5)=(-1)^{16}(40)=40$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(-1)^{1+1}(3(1)-5)=(-1)^2(-2)=-2$ $a_{2}=(-1)^{2+1}(3(2)-5)=(-1)^3(1)=-1$ $a_{3}=(-1)^{3+1}(3(3)-5)=(-1)^4(4)=4$ $a_{4}=(-1)^{4+1}(3(4)-5)=(-1)^5(7)=-7$ $a_{10}=(-1)^{10+1}(3(10)-5)=(-1)^{11}(25)=-25$ $a_{15}=(-1)^{15+1}(3(15)-5)=(-1)^{16}(40)=40$
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