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: 49

Answer

$a_n=(-1)^n(n^2)$

Work Step by Step

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