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: 33

Answer

$a_{1}=(-1)^1((1)^3-1)=0$ $a_{2}=(-1)^2((2)^3-1)=8-1=7$ $a_{3}=(-1)^3((3)^3-1)=-(27-1)=-26$ $a_{4}=(-1)^4((4)^3-1)=64-1=63$ $a_{10}=(-1)^{10}((10)^3-1)=1000-1=999$ $a_{15}=(-1)^{15}((15)^3-1)=-(3375-1)=-3374$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(-1)^1((1)^3-1)=0$ $a_{2}=(-1)^2((2)^3-1)=8-1=7$ $a_{3}=(-1)^3((3)^3-1)=-(27-1)=-26$ $a_{4}=(-1)^4((4)^3-1)=64-1=63$ $a_{10}=(-1)^{10}((10)^3-1)=1000-1=999$ $a_{15}=(-1)^{15}((15)^3-1)=-(3375-1)=-3374$
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