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Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Hello I need help with this question. I tried (3k − 2 is odd by the inductive hypothesis) as alone answer and I got it wrong too.

Transcribed Image Text:A proof by mathematical induction is supposed to show that a given property is true for every integer greater than or equal to an initial value. In order for it to be valid, the property must be
true for the initial value, and the argument in the inductive step must be correct for every integer greater than or equal to the initial value.
Consider the following statement.
For every integer n > 1,
3"
2 is even.
The following is a proposed proof by mathematical induction for the statement.
Since the property is true for n =
1, the basis step is true. Suppose the
property is true for an integer k, where k > 1. That is, suppose that 3* – 2
is even. We must show that 3K + 1
- 2 is even. Observe that
3k + 1
3k. 3 - 2 = 3K(1 + 2) – 2
(3k – 2) + 3k · 2.
2 =
%3D
Now 3K – 2 is even by inductive hypothesis, and 3K· 2 is even by
inspection. Hence the sum of the two quantities is even by
(Theorem 4.1.1). It follows that 3k + 1
2 is even, which is what we
needed to show.
Identify the error(s) in the proof. (Select all that apply.)
The property is not true for n = 1.
3к + 1
2 + (3k – 2) + 3k . 2
The inductive hypothesis is assumed to be true.
(3k – 2) + 3k . 2 + 3k(1 + 2) – 2
3K - 2 is odd by the inductive hypothesis.
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