Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 10: Infinite Sequences and Series - Section 10.1 - Sequences - Exercises 10.1 - Page 569: 5

Answer

$$\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}$$

Work Step by Step

$a_{n}=\frac{2^{n}}{2^{n+1}}=\frac{2^{n}}{2\cdot 2^{n}}=\frac{1}{2},$ so $$a_{1}=a_{2}=a_{3}=a_{4}=\frac{1}{2}$$
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