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: 2

Answer

for all integers n ≥ 1 (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/n) = n + 1

Work Step by Step

let P(n) = all integers ≥ 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/n) = n + 1 Basis step: Show that P(1) is true: P(1) = (1 + 1/1) = 2 by def also P(1) = n + 1 = 1+1 = 2, so P(1) is true. Inductive step: Show that for all integers k ≥ 1, if P(k) is true, then P(k+1) is true: suppose P(k) = k+1is true (inductive hypothesis) P(k+1) = (k+1) + 1 = k+2 (we need to get this result) P(k+1) = 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/k) (1 + 1/(k+1) but 1, (1 + 1/1) (1+1/2) (1+1/3) .....(1+1/k) = P(k) = k+1 so P(k+1) = (k+1) (1+1/(k+1) = (k + 1) + (k+1)/(k+1) = k + 2 [which is what we need to show] Therefore, P(n) is true for all integers n ≥ 1.
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