Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 8 - Sequences and Infinite Series - 8.1 An Overview - 8.1 Exercises - Page 604: 13

Answer

$\frac{4}{3},\frac{8}{5},\frac{16}{9},\frac{32}{17}$

Work Step by Step

$a_n = \frac{2^{n+1}}{2^n + 1}$ To find the first four terms, we need to find $a_1, a_2, a_3$, and $a_4$. $a_1 = \frac{2^{1+1}}{2^1 + 1} = \frac{4}{3}$ $a_2 = \frac{2^{2+1}}{2^2 + 1} = \frac{8}{5}$ $a_3 = \frac{2^{3+1}}{2^3 + 1} = \frac{16}{9}$ $a_4 = \frac{2^{4+1}}{2^4 + 1} = \frac{32}{17}$
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