Arrange the given steps in the correct order to prove that 3" 6, 3k E search e
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Please help me with this two questions. I am having trouble understanding why the answers are incorrect and what the correct answer is. Please provide the correct answer to the following 2 questions below
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![Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction.
Rank the options below.
Suppose that for some k> 6, 3k <k!
3k+1 = 3.3k
3k+1 < (k+1).k!
For n=7, 37=2187 < 7! = 5040.
3k+1 < (k+1)!
3k+1 < (k+1).3k
to search
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Transcribed Image Text:Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction.
Rank the options below.
Suppose that for some k> 6, 3k <k!
3k+1 = 3.3k
3k+1 < (k+1).k!
For n=7, 37=2187 < 7! = 5040.
3k+1 < (k+1)!
3k+1 < (k+1).3k
to search
X
4
X
2
6
3
X
5
X
< Prev
14 of 16
Next >
E
![Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical
induction.
BASIS STEP:
31 (13+2 1), i.e., 3 | 3, so the basis step is true.
INDUCTIVE STEP:
Suppose that 3 | (k³ + 2k).
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + 1).
As the sum of two multiples of 3 is again divisible
by 3, 3 | ((k+1)3 + 2(k + 1)).
31 (03+2 0), i.e., 3 | 0, so the basis step is true.
(k+1)3 + 2(k + 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ +
2k) + 3(k² + 1)
(k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ +
2k) + 3(k² + 1)
Suppose that 3 | (k³ + 2k).
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + k + 1).
As the sum of two multiples of 3 is again divisible
by 3, 3 | ((k+1)3 + 2(k + 1)).
(k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) =
(k3 + 2k)+3(k² + k + 1)
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + k + 1).
31 (13+2 1), i.e., 3 | 3, so the basis step is true.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa1c25850-e50e-4742-88bb-4c5ef336a9a0%2Faaf0807f-1331-4604-8d59-0bda73a29e80%2Fx6pf5vi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical
induction.
BASIS STEP:
31 (13+2 1), i.e., 3 | 3, so the basis step is true.
INDUCTIVE STEP:
Suppose that 3 | (k³ + 2k).
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + 1).
As the sum of two multiples of 3 is again divisible
by 3, 3 | ((k+1)3 + 2(k + 1)).
31 (03+2 0), i.e., 3 | 0, so the basis step is true.
(k+1)3 + 2(k + 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ +
2k) + 3(k² + 1)
(k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ +
2k) + 3(k² + 1)
Suppose that 3 | (k³ + 2k).
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + k + 1).
As the sum of two multiples of 3 is again divisible
by 3, 3 | ((k+1)3 + 2(k + 1)).
(k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) =
(k3 + 2k)+3(k² + k + 1)
By the inductive hypothesis, 3 | (k³ + 2k), and
certainly 3 | 3(k² + k + 1).
31 (13+2 1), i.e., 3 | 3, so the basis step is true.
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