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: 30

Answer

a) $2$ and $1$ b) $a_{n+1} = \frac{1}{2}a_n$, $a_1 = 64$ c) $a_n = 64\times (\frac{1}{2})^{n-1}$

Work Step by Step

a) As we progress through the sequence, the next term is the previous term halved. Thus, the next two terms are $2$ and $1$. b) The first term is $64$, so $a_1 = 64$ The next term is the previous term halved: $a_{n+1} = \frac{1}{2}a_n$ c) The term is $64$ divided by two, $n-1$ times. $a_n = 64\times (\frac{1}{2})^{n-1}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.