The following lemma may be used in this exercise. Lemma: If m is any integer, then m² + m is even. Proof: In case m is odd, then m² is odd and so m² Consider the following statement. For every integer n ≥ 0, n(n² + 5) is divisible by 6. Prove the statement by filling in the blanks below. + m is a sum of two odd integers, which is even. In case m is even, then m² is even and so m² + m is a sum of two even integers, which is even. By the parity principle, m is either even or odd, and in both cases m² + m is even. Proof (by mathematical induction): Let the property P(n) be the following sentence. The expression n(n² + 5) is divisible by 6 Show that P k × is true: P k+1 ☑Your answer cannot be understood or graded. More Information is the following sentence. The expression k(+5) X is divisible by 6 This is true because K(12+5) = 0 , which is divisible by 6 Show that for each integer k ≥ 6 X, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 6 , and suppose that k(+5) is divisible by 6 . (This is the inductive hypothesis.) By definition of divisibility, this implies that there is an integer r such that k³ + 5k = 6 )r, or, equivalently, k³ = 6 r - 5k. In the inductive step, we must show that P(k + 1) is true. In other words, we must show that (k+1)((k+1)²+5) Now is divisible by 6 (k+1)((k+1)² + 5) = (k + 1)((k² + 2 )k + 1) + 5) ( = (k + 1)(k² + ( 2 )k+6) =k³+ (3)k²+(8)k+6 = ((6)r-5k) + 3k²+(8)k+6 by inductive hypothesis we know that there is an integer r such that k³ = 6 r + 3k² + 3 )k + 6 = 6 + 3 ) (k² + k) + 6 (Equation 1). By the lemma, k² + k is even, and so there is an integer s with k² + k = 2s. Substituting into Equation 1 gives that (k+1)((k+1)² + 5) = 6 = 6 )r+ (3 ). ) (r + s + 1) • 2s +6 = 6 )r-5k] By definition of divisibility and the fact that r+ s + 1 is an integer, we can conclude that (k+1)((k+ 1)² + 5) Therefore, P(k + 1) is true. is divisible by 6
The following lemma may be used in this exercise. Lemma: If m is any integer, then m² + m is even. Proof: In case m is odd, then m² is odd and so m² Consider the following statement. For every integer n ≥ 0, n(n² + 5) is divisible by 6. Prove the statement by filling in the blanks below. + m is a sum of two odd integers, which is even. In case m is even, then m² is even and so m² + m is a sum of two even integers, which is even. By the parity principle, m is either even or odd, and in both cases m² + m is even. Proof (by mathematical induction): Let the property P(n) be the following sentence. The expression n(n² + 5) is divisible by 6 Show that P k × is true: P k+1 ☑Your answer cannot be understood or graded. More Information is the following sentence. The expression k(+5) X is divisible by 6 This is true because K(12+5) = 0 , which is divisible by 6 Show that for each integer k ≥ 6 X, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 6 , and suppose that k(+5) is divisible by 6 . (This is the inductive hypothesis.) By definition of divisibility, this implies that there is an integer r such that k³ + 5k = 6 )r, or, equivalently, k³ = 6 r - 5k. In the inductive step, we must show that P(k + 1) is true. In other words, we must show that (k+1)((k+1)²+5) Now is divisible by 6 (k+1)((k+1)² + 5) = (k + 1)((k² + 2 )k + 1) + 5) ( = (k + 1)(k² + ( 2 )k+6) =k³+ (3)k²+(8)k+6 = ((6)r-5k) + 3k²+(8)k+6 by inductive hypothesis we know that there is an integer r such that k³ = 6 r + 3k² + 3 )k + 6 = 6 + 3 ) (k² + k) + 6 (Equation 1). By the lemma, k² + k is even, and so there is an integer s with k² + k = 2s. Substituting into Equation 1 gives that (k+1)((k+1)² + 5) = 6 = 6 )r+ (3 ). ) (r + s + 1) • 2s +6 = 6 )r-5k] By definition of divisibility and the fact that r+ s + 1 is an integer, we can conclude that (k+1)((k+ 1)² + 5) Therefore, P(k + 1) is true. is divisible by 6
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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Question
need help only answer the incorrect answers
![The following lemma may be used in this exercise.
Lemma: If m is any integer, then m² + m is even.
Proof: In case m is odd, then m² is odd and so m²
Consider the following statement.
For every integer n ≥ 0, n(n² + 5) is divisible by 6.
Prove the statement by filling in the blanks below.
+ m is a sum of two odd integers, which is even. In case m is even, then m² is even and so m² + m is a sum of two even integers, which is even. By the parity principle, m is either even or odd, and in both cases m² + m is even.
Proof (by mathematical induction): Let the property P(n) be the following sentence.
The expression n(n² + 5)
is divisible by 6
Show that P k ×
is true: P
k+1 ☑Your answer cannot be understood or graded. More Information is the following sentence.
The expression k(+5)
X is divisible by 6
This is true because K(12+5)
= 0
, which is divisible by 6
Show that for each integer k ≥ 6
X, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k ≥ 6
, and suppose that k(+5)
is divisible by 6
. (This is the inductive hypothesis.)
By definition of divisibility, this implies that there is an integer r such that k³ + 5k =
6
)r,
or, equivalently, k³ =
6
r - 5k.
In the inductive step, we must show that P(k + 1) is true. In other words, we must show that (k+1)((k+1)²+5)
Now
is divisible by 6
(k+1)((k+1)² + 5)
=
(k + 1)((k²
+ 2
)k + 1) + 5)
(
= (k + 1)(k² + ( 2 )k+6)
=k³+ (3)k²+(8)k+6
= ((6)r-5k) + 3k²+(8)k+6
by inductive hypothesis we know that there is an integer r such that k³
=
6
r + 3k² +
3
)k + 6
=
6
+ 3
) (k²
+ k) + 6
(Equation 1).
By the lemma, k² + k is even, and so there is an integer s with k² + k = 2s.
Substituting into Equation 1 gives that
(k+1)((k+1)² + 5) =
6
=
6
)r+ (3
).
) (r + s + 1)
• 2s +6
= 6
)r-5k]
By definition of divisibility and the fact that r+ s + 1 is an integer, we can conclude that (k+1)((k+ 1)² + 5)
Therefore, P(k + 1) is true.
is divisible by 6](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3cf874b-7a7b-478f-a7b8-421442e72224%2Fcbbc89b3-3a22-4aeb-a10c-76cc1339a035%2Fgfvumq_processed.png&w=3840&q=75)
Transcribed Image Text:The following lemma may be used in this exercise.
Lemma: If m is any integer, then m² + m is even.
Proof: In case m is odd, then m² is odd and so m²
Consider the following statement.
For every integer n ≥ 0, n(n² + 5) is divisible by 6.
Prove the statement by filling in the blanks below.
+ m is a sum of two odd integers, which is even. In case m is even, then m² is even and so m² + m is a sum of two even integers, which is even. By the parity principle, m is either even or odd, and in both cases m² + m is even.
Proof (by mathematical induction): Let the property P(n) be the following sentence.
The expression n(n² + 5)
is divisible by 6
Show that P k ×
is true: P
k+1 ☑Your answer cannot be understood or graded. More Information is the following sentence.
The expression k(+5)
X is divisible by 6
This is true because K(12+5)
= 0
, which is divisible by 6
Show that for each integer k ≥ 6
X, if P(k) is true, then P(k + 1) is true:
Let k be any integer with k ≥ 6
, and suppose that k(+5)
is divisible by 6
. (This is the inductive hypothesis.)
By definition of divisibility, this implies that there is an integer r such that k³ + 5k =
6
)r,
or, equivalently, k³ =
6
r - 5k.
In the inductive step, we must show that P(k + 1) is true. In other words, we must show that (k+1)((k+1)²+5)
Now
is divisible by 6
(k+1)((k+1)² + 5)
=
(k + 1)((k²
+ 2
)k + 1) + 5)
(
= (k + 1)(k² + ( 2 )k+6)
=k³+ (3)k²+(8)k+6
= ((6)r-5k) + 3k²+(8)k+6
by inductive hypothesis we know that there is an integer r such that k³
=
6
r + 3k² +
3
)k + 6
=
6
+ 3
) (k²
+ k) + 6
(Equation 1).
By the lemma, k² + k is even, and so there is an integer s with k² + k = 2s.
Substituting into Equation 1 gives that
(k+1)((k+1)² + 5) =
6
=
6
)r+ (3
).
) (r + s + 1)
• 2s +6
= 6
)r-5k]
By definition of divisibility and the fact that r+ s + 1 is an integer, we can conclude that (k+1)((k+ 1)² + 5)
Therefore, P(k + 1) is true.
is divisible by 6
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