Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.2 - Mathematical Induction - Exercises A.2 - Page AP-9: 8


See proof below.

Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=-3$ $LHS=2^{-3}=\displaystyle \frac{1}{2^{3}}=\frac{1}{8}$ so, $LHS\displaystyle \geq\frac{1}{8}\quad $ when $n=-3.$ Yes, this is true when $n=-3.$ $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k\geq-3$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k\geq-3$, $ 2^{k}\geq \displaystyle \frac{1}{8}$ (we call this the step hypothesis) For $n=k+1$, the LHS equals$\quad 2^{k+1}$ which we rewrite as $LHS= 2^{k+1}=(2^{k})\cdot 2\quad $ and we apply the hypothesis $ 2^{k+1}\displaystyle \geq\frac{1}{8}\cdot 2$ Now, we apply the fact that 2 is greater than 1 $2^{k+1}\displaystyle \geq\frac{1}{8}\cdot 1=\frac{1}{8}$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n\geq-3.$
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