(Discrete Math HW)Please Check mark the correct boxesthank you 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 forthe 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 2 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 theproperty is true for an integer k, where k > 1. That is, suppose that3K – 2 is even. We must show that 3K + 1 - 2 is even. Observe that%3D3k. 3 – 2 = 3k(1 + 2) – 2(3k – 2) + 3k . 2.3k + 12Now 3K – 2 is even by inductive hypothesis, and 3K · 2 is even byinspection. Hence the sum of the two quantities is even by(Theorem 4.1.1). It follows that 3* + - - 2 is even, which is what weneeded to show.Identify the error(s) in the proof. (Select all that apply.)3k + 1- 2 + (3k – 2) + 3k . 2The inductive hypothesis is assumed to be true.The property is not true for n = 1.3k2 is odd by the inductive hypothesis.O (3k – 2) + 3k. 2 + 3*(1 + 2) – 2

It is given to be proved that, 3n-2 is even for n≥1. But for n = 1 we see that, 31-2=3-2=1; (not an…

Answered: following statement. For every integer…

following statement. For every integer n 2 1, 3" - 2 is even. g is a proposed proof by mathematical induction for the statement.

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter8: Sequences, Series, And Probability

Section8.5: Mathematical Induction

Problem 42E

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Question

(Discrete Math HW)

Please Check mark the correct boxes

thank you

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 2 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
3K – 2 is even. We must show that 3K + 1 - 2 is even. Observe that
%3D
3k. 3 – 2 = 3k(1 + 2) – 2
(3k – 2) + 3k . 2.
3k + 1
2
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 3* + - - 2 is even, which is what we
needed to show.
Identify the error(s) in the proof. (Select all that apply.)
3k + 1
- 2 + (3k – 2) + 3k . 2
The inductive hypothesis is assumed to be true.
The property is not true for n = 1.
3k
2 is odd by the inductive hypothesis.
O (3k – 2) + 3k. 2 + 3*(1 + 2) – 2

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