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:
$a_n=(-1)^n(n^2)$
Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
by Bittinger, Marvin L.; Ellenbogen, David J.; Johnson, Barbara L.
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 - Page 894: 49
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