Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.1 Sequences - 11.1 Exercises - Page 744: 17


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

Work Step by Step

Our first term is $a_1 = \frac{1}{2}$. The sequence alternates, so we have a factor of $(-1)^n$. We observe that we are also multiplying by $\frac{n^2}{n+1}$ each term. Since $a_1 = \frac{1}{2}$, we must have $$a_n = -\frac{(-1)^nn^2}{n+1}.$$
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