University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 9 - Section 9.1 - Sequences - Exercises - Page 487: 5

Answer

$a_1=a_2=a_3=a_4=\frac{1}{2}$

Work Step by Step

We know that $a_n=\frac{2^n}{2^{n+1}}=\frac{2^n}{2\cdot2^n}=\frac{1}{2}$. Hence, here: $a_1=a_2=a_3=a_4=\frac{1}{2}$
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