5 =5 Hence, 6"-1 is divisible by 5 is (10.). for n=1 Part 2: Assume that 6" - 1 is divisible by 5 for n-k21 6* -1 is divisible by 5 6*+1 -1 is divisible by 5 6*+1 -1= (11.). (12.). = [(6 – 1)- 6*]+ (6* - 1) = %3D %3D By divisibility definition, 5. 6* and 6* - 1 are divisible by 5. Therefore, their sum is also divisible by 5. The given divisibility statement, 6" - 1 is divisible by 5, has satisfied the two conditions of the (13.) positive integers n Hence, 6" - 1 is divisible by 5 for all
5 =5 Hence, 6"-1 is divisible by 5 is (10.). for n=1 Part 2: Assume that 6" - 1 is divisible by 5 for n-k21 6* -1 is divisible by 5 6*+1 -1 is divisible by 5 6*+1 -1= (11.). (12.). = [(6 – 1)- 6*]+ (6* - 1) = %3D %3D By divisibility definition, 5. 6* and 6* - 1 are divisible by 5. Therefore, their sum is also divisible by 5. The given divisibility statement, 6" - 1 is divisible by 5, has satisfied the two conditions of the (13.) positive integers n Hence, 6" - 1 is divisible by 5 for all
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.4: Fractional Expressions
Problem 68E
Related questions
Question
![5 = 5
Hence, 6" - 1 is divisible by 5 is (10.)_
for n=1
Part 2: Assume that 6" - 1 is divisible by 5 for n k21
6* - 1 is divisible by 5
6*+1 -1 is divisible by 5
6k*1 -1 = (11.) _= [(6-1) 64]+ (6 - 1) =
(12.)
By divisibility definition, 5 6k and 6* - 1 are divisible by 5. Therefore,
their sum is also divisible by 5.
The given divisibility statement, 6" -1 is divisible by 5, has satisfied the two
conditions of the (13.)
positive integers n
.Hence, 6"-1 is divisible by 5 for all
3. 2" > 2n for every integer n 23
Solution:
Part 1: The claim of the statement, 2" > 2n for every integer n 2 3. Hence,
n-(14.)
2 > 2n = (15.).
Therefore, the statement is true when (16.)
= 8>6
Part 2: Assume that 2" > 2n for every integer n 2 3 for n-k 21
2* > 2k can be written as (17.)_
inequality
by Multiplication property of
2k+1 > 2(k + 1)
2k - 21 > 2(k + 1)
(18.)
The inequality is (19.)
illustrates the necessity of Part 1 of the proof to establish the result. However, the result
above can be modified to 2" > 2n for all nonnegative integers n
for n 23 and (20.)
for n <3. This
matical
tive](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F99eb080c-44ec-40da-b239-326c83448536%2Fa3dab2b5-21b7-4c51-aa32-35cdf83e485d%2Fw2fxf9ve_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5 = 5
Hence, 6" - 1 is divisible by 5 is (10.)_
for n=1
Part 2: Assume that 6" - 1 is divisible by 5 for n k21
6* - 1 is divisible by 5
6*+1 -1 is divisible by 5
6k*1 -1 = (11.) _= [(6-1) 64]+ (6 - 1) =
(12.)
By divisibility definition, 5 6k and 6* - 1 are divisible by 5. Therefore,
their sum is also divisible by 5.
The given divisibility statement, 6" -1 is divisible by 5, has satisfied the two
conditions of the (13.)
positive integers n
.Hence, 6"-1 is divisible by 5 for all
3. 2" > 2n for every integer n 23
Solution:
Part 1: The claim of the statement, 2" > 2n for every integer n 2 3. Hence,
n-(14.)
2 > 2n = (15.).
Therefore, the statement is true when (16.)
= 8>6
Part 2: Assume that 2" > 2n for every integer n 2 3 for n-k 21
2* > 2k can be written as (17.)_
inequality
by Multiplication property of
2k+1 > 2(k + 1)
2k - 21 > 2(k + 1)
(18.)
The inequality is (19.)
illustrates the necessity of Part 1 of the proof to establish the result. However, the result
above can be modified to 2" > 2n for all nonnegative integers n
for n 23 and (20.)
for n <3. This
matical
tive
Transcribed Image Text:2. 6" - 1 is divisible by 5
Solution:
Part 1: Prove that 6"-1 is divisible by 5 is true for n-1
(9.)
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