Fill in the blanks in the proof of the of fact that 3" 1 is a multiple of 2 for every natural n. Base case: if k= [ Select] then 31 1= 2 is a multiple of 2, as needed. Induction step: Assume that for [ Select] v natural k, 3k 1 is a multiple of 2. Consider [ Select] v Then n= 3k+1 -1= 3k.3-1= 3* (2+ 1) - 1 = 3k - 2 - (3k 1). %3D Notice that the first term is a multiple of 2 and the second term is a multiple of 2 by induction assumption. Therefore, 3k+1 - 1 is a multiple of 2, as required. The proof is complete.
Fill in the blanks in the proof of the of fact that 3" 1 is a multiple of 2 for every natural n. Base case: if k= [ Select] then 31 1= 2 is a multiple of 2, as needed. Induction step: Assume that for [ Select] v natural k, 3k 1 is a multiple of 2. Consider [ Select] v Then n= 3k+1 -1= 3k.3-1= 3* (2+ 1) - 1 = 3k - 2 - (3k 1). %3D Notice that the first term is a multiple of 2 and the second term is a multiple of 2 by induction assumption. Therefore, 3k+1 - 1 is a multiple of 2, as required. The proof is complete.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Selection 1: 0, 1 , 2
Selection 2: some, any
Selection 3: k+2, k, k+1, k-1
![Fill in the blanks in the proof of the of fact that 3"
1 is a multiple of 2 for every natural n.
"Base case: if k= [Select]
then 31
1= 2 is a multiple of 2, as needed.
Induction step: Assume that for [Select]
natural k, 3k
1 is a multiple of 2. Consider
[ Select ]
Then
n=
3k+1 – 1= 3*. 3 -1= 3*(2+ 1) –1 = 3* . 2 - (3k 1).
Notice that the first term is a multiple of 2 and the second term is a multiple of 2 by induction assumption.
Therefore, 3k+1
1 is a multiple of 2, as required. The proof is complete.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5f26210-6cc2-465b-a880-67054eb449ea%2F234273ff-a29c-48a7-bb7d-ff0905c8f382%2Fujqrzji_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Fill in the blanks in the proof of the of fact that 3"
1 is a multiple of 2 for every natural n.
"Base case: if k= [Select]
then 31
1= 2 is a multiple of 2, as needed.
Induction step: Assume that for [Select]
natural k, 3k
1 is a multiple of 2. Consider
[ Select ]
Then
n=
3k+1 – 1= 3*. 3 -1= 3*(2+ 1) –1 = 3* . 2 - (3k 1).
Notice that the first term is a multiple of 2 and the second term is a multiple of 2 by induction assumption.
Therefore, 3k+1
1 is a multiple of 2, as required. The proof is complete.
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