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: 2


See the proof below.

Work Step by Step

$1. \ (\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e})$ Check that the formula holds for n = 1. LHS$=1+r^{1},\qquad $ RHS$=\displaystyle \frac{1-r^{1+1}}{1+r}=\frac{1-r^{2}}{1-r}=\frac{(1-r)(1+r)}{1-r}$ We recognized a difference of squares in the numerator. We now cancel the common factor, $RHS=1+r=LHS\qquad $... yes, it 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 $n=k$, $1+r^{1}+r^{2}+...+r^{k}=\displaystyle \frac{1-r^{k+1}}{1-r}$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $1+r^{1}+r^{2}+...+r^{k}+r^{k+1}$ which we rewrite as $LHS=(1+r^{1}+r^{2}+...+r^{k})+r^{k+1}=\qquad $ use the hypothesis: $=\displaystyle \frac{1-r^{k+1}}{1-r}+r^{k+1}$ $=\displaystyle \frac{1-r^{k+1}+r^{k+1}(1-r)}{1-r}$ $=\displaystyle \frac{1-r^{k+1}+r^{k+1}-r^{k+2}}{1-r}$ $=\displaystyle \frac{1-r^{k+2}}{1-r}$ $=\displaystyle \frac{1-r^{(k+1)+1}}{1-r}=RHS,\quad $ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n.$
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