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: 50

Answer

$a_n=(-1)^{n+1}(n^2)$

Work Step by Step

If we look at the pattern: $1, -4, 9, -16, ...$ $(-1)^2(1^2), (-1)^3(2^2), (-1)^4(3^2), (-1)^5(4^2) ...$ The nth term in the sequence is: $a_n=(-1)^{n+1}(n^2)$
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