Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 11 - Section 11.1 - Sequences - 11.1 Exercises - Page 704: 17

Answer

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

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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