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By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = 10" 1 is a solution to the recurrence relation an 10an-1 + 9 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a ₁ 10ao + 9. Then 10k+1 −1= ak+1 10(10k – 1) + 9, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. By the recurrence relation, we have ak+1 = 10ak + 9 = 10(10¹ − 1) + 9 - Thus P(k+ 1) is true. Now assume that P(k + 1) is true. This simplifies to 10+¹ − 10 + 9 = 10k+1 . – 1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the statement, "an = 10" 1 is a solution to the recurrence relation an = 10an-1 + 9 with a = 0." Now assume that P(n) is true for all n ≥ 0. 10⁰ 11-1 = 0, as Note that a required. = Now assume that P(k) is true for an arbitrary integer k ≥ 1. 10k - 1. Let P(n) be the statement, "an Then ak+1 = Then = 10" 1". 10ak + 9, so P(k + 1) is true. Thus P(k) is true for all k.

By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = 10" 1 is a solution to the recurrence relation an 10an-1 + 9 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a ₁ 10ao + 9. Then 10k+1 −1= ak+1 10(10k – 1) + 9, which is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. By the recurrence relation, we have ak+1 = 10ak + 9 = 10(10¹ − 1) + 9 - Thus P(k+ 1) is true. Now assume that P(k + 1) is true. This simplifies to 10+¹ − 10 + 9 = 10k+1 . – 1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the statement, "an = 10" 1 is a solution to the recurrence relation an = 10an-1 + 9 with a = 0." Now assume that P(n) is true for all n ≥ 0. 10⁰ 11-1 = 0, as Note that a required. = Now assume that P(k) is true for an arbitrary integer k ≥ 1. 10k - 1. Let P(n) be the statement, "an Then ak+1 = Then = 10" 1". 10ak + 9, so P(k + 1) is true. Thus P(k) is true for all k.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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please help and please use the wording in the given statements thank you!

By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = 10" - 1 is a solution to the recurrence relation an = 10an-1 + 9 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: 10ao + 9. Note that a₁ Then 10k+1 1 = 10(10k − 1) + 9, which is true. = Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. ak+1 By the recurrence relation, we have ak+1 10ak + 9 = 10(10k − 1) +9 = Thus P(k+ 1) is true. Now assume that P(k + 1) is true. This simplifies to = 10k+1 − 10 + 9 = 10k+1 = - - 1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the statement, "an = 10" 1 is a solution to the recurrence relation 10an-1 + 9 with a = 0." Now assume that P(n) is true for all n ≥ 0. 10⁰ 11-1 = 0, as required. an = Note that a = Now assume that P(k) is true for an arbitrary integer k > 1. Then = 10¹ - 1. Let P(n) be the statement, "an Then ak+1= = 10" 1". 10ak + 9, so P(k + 1) is true. Thus P(k) is true for all k.
Transcribed Image Text:By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = 10" - 1 is a solution to the recurrence relation an = 10an-1 + 9 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: 10ao + 9. Note that a₁ Then 10k+1 1 = 10(10k − 1) + 9, which is true. = Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. ak+1 By the recurrence relation, we have ak+1 10ak + 9 = 10(10k − 1) +9 = Thus P(k+ 1) is true. Now assume that P(k + 1) is true. This simplifies to = 10k+1 − 10 + 9 = 10k+1 = - - 1 Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the statement, "an = 10" 1 is a solution to the recurrence relation 10an-1 + 9 with a = 0." Now assume that P(n) is true for all n ≥ 0. 10⁰ 11-1 = 0, as required. an = Note that a = Now assume that P(k) is true for an arbitrary integer k > 1. Then = 10¹ - 1. Let P(n) be the statement, "an Then ak+1= = 10" 1". 10ak + 9, so P(k + 1) is true. Thus P(k) is true for all k.
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