Discrete Mathematics with Applications 4th Edition

Published by Cengage Learning
ISBN 10: 0-49539-132-8
ISBN 13: 978-0-49539-132-6

Chapter 5 - Sequences, Mathematical Induction, and Recursion - Exercise Set 5.3 - Page 266: 1

Answer

$\frac{1}{k+1}$

Work Step by Step

General formula : $\pi_{i=2}^n(1-\frac{1}{i})=\frac{1}{n}$ for all integers $n\geq 2$ Proof (by mathematical induction) let the property p(n) by the equation $\pi^n_{i=2}(1-\frac{1}{i})=\frac{1}{n}$ Show that P(2) is true: The left-hand side of P(2) is $\pi^n_{i=2}(1-\frac{1}{i})=\frac{1}{n}=1-\frac{1}{2}=\frac{1}{2}$ which equals the right-hand side. Show that for all integers k ≥ 2, if P(k) is true then P(k + 1) is also true: Suppose that k is any integer with k ≥ 2 such that $\pi^k_{i=2}(1-\frac{1}{i})=\frac{1}{k}$ We must show that $\pi^{k+1}_{i=2}(1-\frac{1}{i})=\frac{1}{k+1}$ But by the laws of algebra and substitution from the inductive hypothesis, the left-hand side of P(k + 1) is $\pi^{k+1}_{i=2}(1-\frac{1}{i})$ =$\pi^{k}_{i=2}(1-\frac{1}{i})(1-\frac{1}{k+1})$ =$(\frac{1}{k})(1-\frac{1}{k+1})=(\frac{1}{k})(\frac{(k+1)-1}{k+1})$ =$\frac{1}{k+1}$ which is the right-hand side of P(k + 1)
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