Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 11 - Sequences; Induction; the Binomial Theorem - Section 11.4 Mathematical Induction - 11.4 Assess Your Understanding - Page 850: 27

Answer

see below.

Work Step by Step

1. Test for $n=1$, $1+a\ge 1+a$, it is true. 2. Assume the statement is true for $n=k$, that is $(1+a)^k\ge1+ka$, 3. For $n=k+1$, with $a\gt0$, we have $(1+a)^{k+1}=(1+a)(1+a)^k\ge(1+a)(1+ka)=1+ka+a+ka^2=1+(k+1)a+ka^2\ge1+(k+1)a$, thus it is true for $n=k+1$, 4. We can conclude that the statement is true for any integer $n$.
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