By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = 7 - 1 is a solution to the recurrence relation an = 7an-1 +6 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: Your Proof: Put chosen statements in order in this

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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Then 7k+1 - 1 = 7(7k . − 1) + 6, which is true.
Now assume that P(k + 1) is true.
This simplifies to
=
: 7k+¹ − 7 + 6 = 7k+1
-
Thus P(k+ 1) is true.
Therefore, by the Principle of Mathematical
Induction, P(n) is true for all n ≥ 1.
Now assume that P(n) is true for all n ≥ 0.
Let P(n) be the statement, "an = 7" - 1 is a
solution to the recurrence relation
an = 7an-1 +6 with a = 0."
Thus P(k) is true for all k.
Then = 7¹ - 1.
ak+1
1
By the recurrence relation, we have
ak+1 = 7ak + 6 = 7(7⁰ − 1) + 6
Transcribed Image Text:Then 7k+1 - 1 = 7(7k . − 1) + 6, which is true. Now assume that P(k + 1) is true. This simplifies to = : 7k+¹ − 7 + 6 = 7k+1 - Thus P(k+ 1) is true. Therefore, by the Principle of Mathematical Induction, P(n) is true for all n ≥ 1. Now assume that P(n) is true for all n ≥ 0. Let P(n) be the statement, "an = 7" - 1 is a solution to the recurrence relation an = 7an-1 +6 with a = 0." Thus P(k) is true for all k. Then = 7¹ - 1. ak+1 1 By the recurrence relation, we have ak+1 = 7ak + 6 = 7(7⁰ − 1) + 6
By dragging statements from the left column to the right column below, give a proof by induction of the following statement:
an7" 1 is a solution to the recurrence relation an = 7an-1 +6 with a = 0.
The correct proof will use 8 of the statements below.
Statements to choose from:
Note that a = 7⁰ -1 1-1 = 0, as
required.
Let P(n) be the statement, "an = 7" – 1".
Now assume that P(k) is true for an arbitrary
integer k ≥ 1.
Then ak+1
Then 7k+1
=
7ak + 6, so P(k + 1) is true.
7a0 + 6.
−1= 7(7k – 1) + 6, which is true.
Note that a₁ =
Now assume that P(k + 1) is true.
This simplifies to
: 7k+¹ − 7 + 6 = 7k+1 _ 1
ak+1 =
Your Proof: Put chosen statements in order in this
column and press the Submit Answers button.
Transcribed Image Text:By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an7" 1 is a solution to the recurrence relation an = 7an-1 +6 with a = 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a = 7⁰ -1 1-1 = 0, as required. Let P(n) be the statement, "an = 7" – 1". Now assume that P(k) is true for an arbitrary integer k ≥ 1. Then ak+1 Then 7k+1 = 7ak + 6, so P(k + 1) is true. 7a0 + 6. −1= 7(7k – 1) + 6, which is true. Note that a₁ = Now assume that P(k + 1) is true. This simplifies to : 7k+¹ − 7 + 6 = 7k+1 _ 1 ak+1 = Your Proof: Put chosen statements in order in this column and press the Submit Answers button.
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