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: 28

Answer

a) $36$ and $49$ b) $a_{n+1} = (\sqrt a_n + 1)^2$, $a_1 = 1$ c) $a_n = n^2$

Work Step by Step

a) Another way to write the sequence would be: $\{1^2, 2^2, 3^2, 4^2, 5^2,...\}$. Thus, the next two terms are $6^2$ and $7^2$ or $36$ and $49$. b) The first term is $1$, so $a_1 = 1$ We have to find what value was squared in the previous term, add one, then square the new value to get the next term. $a_{n+1} = (\sqrt a_n + 1)^2$ c) The term is simply the value of $n$ squared. $a_n = n^2$
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