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: 5


See proof below.

Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for $n=2$. We are given by the problem text that $LHS=\displaystyle \frac{2}{3}$ $RHS=1-\displaystyle \frac{1}{3^{1}}=\frac{3-1}{3}=\frac{2}{3}=LHS \qquad $ Yes, this is true when $n=$1. $2. \ (\mathrm{s}\mathrm{t}\mathrm{e}\mathrm{p})$ Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k$, $\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}}=1-\frac{1}{3^{k}}$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}}+\frac{2}{3^{k+1}},\quad $ which we rewrite as $LHS= (\displaystyle \frac{2}{3}+\frac{2}{3^{2}}+...+\frac{2}{3^{k}})+\frac{2}{3^{k+1}}$ apply the hypothesis on the first set of parentheses $=1-\displaystyle \frac{1}{3^{k}}+\frac{2}{3^{k+1}}$ $=1-(\displaystyle \frac{1}{3^{k}}-\frac{2}{3^{k+1}})$ $=1-\displaystyle \frac{3-2}{3^{k+1}}$ $=1-\displaystyle \frac{1}{3^{k+1}}=RHS$ for $n=k+1$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n$.
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