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: 15


$$a_n = (-1)^n\frac{2^{n-1}}{3^{n-2}}.$$

Work Step by Step

Our first term is $a_1 = -3$. Because the sequence alternates, we have a $(-1)^{n}$ in the sequence. Each term also changes by a factor of $\frac{2}{3}$, so we introduce a factor of $(\frac{2}{3})^n$. Putting together the pieces with the fact that $a_1 = -3$, we have: $$a_n = (-1)^n3(\frac{2}{3})^{n-1} = (-1)^n\frac{2^{n-1}}{3^{n-2}}$$
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