Fill in the missing pieces in the following proof that1 +

Discrete Mathematics with Applications | 4th Edition | ISBN: 9780495391326 | Authors: Susanna S. Epp

Problem 5E Chapter 5.2

Discrete Mathematics with Applications | 4th Edition

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Discrete Mathematics with Applications | 4th Edition | ISBN: 9780495391326 | Authors: Susanna S. Epp

Discrete Mathematics with Applications | 4th Edition

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Problem 5E

Fill in the missing pieces in the following proof that

1 + 3 + 5 + … + (2n –1) = n

for all integers n > 1.

Proof: Let the property P (n) be the equation

1 + 3 + 5 + +(2n –1) = n2. ← P(n)

Show that P(1) is true: To establish P (1), we must show that when 1 is substituted in place of n, the left-hand side equals the right-hand side. But when n = 1, the left-hand side is the sum of all the odd integers from 1 to 2•1 – 1, which is the sum of the odd integers from 1 to 1, which is just 1. The right-hand side is (a) which also equals 1. So P(1 ) is true.

Show that for all integers k > 1, if P (k) is true then P (k +1) is true: Let k be any integer with k ≥ 1.

[Suppose P (k) is true. That is:]

Suppose 1 + 3 + 5 + … + (2k – 1) =

[This is the inductive hypothesis.]

[We must show that P (k + 1) is true. That is:]

We must show that

But the left-hand side of P (k + 1) is

which is the right-hand side of P (k + 1) [as was to be shown.]

[Since we have proved the basis step and the inductive step, we conclude that the given statement is true.]Publisher information found

The previous proof was annotated to help make its logical flow more obvious. In standard mathematical writing, such annotation is omitted.

Step-by-Step Solution:
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Chapter 5.2, Problem 5E is Solved
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Textbook: Discrete Mathematics with Applications
Edition: 4th
Author: Susanna S. Epp
ISBN: 9780495391326

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Fill in the missing pieces in the following proof that1 +

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