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: 30

Answer

$a_{1}=(-2)^{1+1}=(-2)^2=4$ $a_{2}=(-2)^{2+1}=(-2)^3=-8$ $a_{3}=(-2)^{3+1}=(-2)^4=16$ $a_{4}=(-2)^{4+1}=(-2)^5=-32$ $a_{10}=(-2)^{10+1}=(-2)^{11}=-2048$ $a_{15}=(-2)^{15+1}=(-2)^{16}=65536$

Work Step by Step

If we want to find a term, we have to substitute $n$ by its index: $a_{1}=(-2)^{1+1}=(-2)^2=4$ $a_{2}=(-2)^{2+1}=(-2)^3=-8$ $a_{3}=(-2)^{3+1}=(-2)^4=16$ $a_{4}=(-2)^{4+1}=(-2)^5=-32$ $a_{10}=(-2)^{10+1}=(-2)^{11}=-2048$ $a_{15}=(-2)^{15+1}=(-2)^{16}=65536$
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