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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9* − 1) + 8, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1.

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9* − 1) + 8, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1.

Operations Research : Applications and Algorithms
Operations Research : Applications and Algorithms
4th Edition
ISBN:9780534380588
Author:Wayne L. Winston
Publisher:Wayne L. Winston
Chapter2: Basic Linear Algebra
Section: Chapter Questions
Problem 15RP
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(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9* − 1) + 8, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1.
Transcribed Image Text:(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9* − 1) + 8, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1.
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