Question B2: Last week, we noted that 2n³ +3n² + n can be factored as the product (2n + 1)(n+1)(n). One approach to show that this is always divisible by 3 is to prove that for every integer n, at least one of the factors is a multiple of 3. By tabulating values, it became clear that the multiple of 3 is not always the same factor, but that it changes with some kind of pattern. As with even and odd numbers, we can classify integers according to their remainder upon division by 3, so that exactly one of the following is always true: 3k(n = 3k) or 3k(n = 3k + 1) or 3k(n = 3k + 2) (a) Show that if n = 3k, then 2n³ + 3n² + n = 3[(2n + 1)(n+1)(k)], which is a multiple of 3. (b) Show that if n = 3k +2, then 2n³ + 3n² + n = 3[(2n + 1)(k+1)(n)], which is a multiple of 3. (c) Find an expression for 2n³ + 3n² +n when n = 3k + 1 that shows the polynomial is a multiple of 3. Together these facts are enough to show (proof by cases) that 2n³ + 3n² +n is always a multiple of 3. (d) Prove that if an integer m is a multiple of 3 and also an even number, then it must be a multiple of 6. Note: When we say a is a multiple of b, we are asserting the claim 3k(a = kb). You need to show the implication [(k(m = 3k)) ^ (3k(m = 2k))] → [3k(m= 6k)]. How are the k's related?
Question B2: Last week, we noted that 2n³ +3n² + n can be factored as the product (2n + 1)(n+1)(n). One approach to show that this is always divisible by 3 is to prove that for every integer n, at least one of the factors is a multiple of 3. By tabulating values, it became clear that the multiple of 3 is not always the same factor, but that it changes with some kind of pattern. As with even and odd numbers, we can classify integers according to their remainder upon division by 3, so that exactly one of the following is always true: 3k(n = 3k) or 3k(n = 3k + 1) or 3k(n = 3k + 2) (a) Show that if n = 3k, then 2n³ + 3n² + n = 3[(2n + 1)(n+1)(k)], which is a multiple of 3. (b) Show that if n = 3k +2, then 2n³ + 3n² + n = 3[(2n + 1)(k+1)(n)], which is a multiple of 3. (c) Find an expression for 2n³ + 3n² +n when n = 3k + 1 that shows the polynomial is a multiple of 3. Together these facts are enough to show (proof by cases) that 2n³ + 3n² +n is always a multiple of 3. (d) Prove that if an integer m is a multiple of 3 and also an even number, then it must be a multiple of 6. Note: When we say a is a multiple of b, we are asserting the claim 3k(a = kb). You need to show the implication [(k(m = 3k)) ^ (3k(m = 2k))] → [3k(m= 6k)]. How are the k's related?
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
discrete math
![Question B2:
Last week, we noted that 2n³ +3n² + n can be factored as the product (2n + 1)(n+1)(n). One approach
to show that this is always divisible by 3 is to prove that for every integer n, at least one of the factors is a
multiple of 3. By tabulating values, it became clear that the multiple of 3 is not always the same factor,
but that it changes with some kind of pattern. As with even and odd numbers, we can classify integers
according to their remainder upon division by 3, so that exactly one of the following is always true:
3k(n = 3k) or 3k(n = 3k + 1)
or 3k(n = 3k + 2)
(a) Show that if n = 3k, then 2n³ + 3n² + n = : 3[(2n + 1)(n+1)(k)], which is a multiple of 3.
(b) Show that if n = 3k + 2, then 2n³ + 3n² +n=3[(2n + 1)(k+ 1)(n)], which is a multiple of 3.
(c) Find an expression for 2n³ +3n² +n when n = 3k + 1 that shows the polynomial is a multiple of 3.
Together these facts are enough to show (proof by cases) that 2n³ + 3n² +n is always a multiple of 3.
(d) Prove that if an integer m is a multiple of 3 and also an even number, then it must be a multiple of 6.
Note: When we say a is a multiple of b, we are asserting the claim 3k(a = kb). You need to show the
implication [(k(m = 3k)) ^ (³k(m = 2k))] → [³k(m = 6k)]. How are the k's related?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7355d962-0fce-4254-b762-05ebe0225361%2F8e52763f-4ef3-4152-b1b2-b32fd5613be1%2F5ad47m2_processed.png&w=3840&q=75)
Transcribed Image Text:Question B2:
Last week, we noted that 2n³ +3n² + n can be factored as the product (2n + 1)(n+1)(n). One approach
to show that this is always divisible by 3 is to prove that for every integer n, at least one of the factors is a
multiple of 3. By tabulating values, it became clear that the multiple of 3 is not always the same factor,
but that it changes with some kind of pattern. As with even and odd numbers, we can classify integers
according to their remainder upon division by 3, so that exactly one of the following is always true:
3k(n = 3k) or 3k(n = 3k + 1)
or 3k(n = 3k + 2)
(a) Show that if n = 3k, then 2n³ + 3n² + n = : 3[(2n + 1)(n+1)(k)], which is a multiple of 3.
(b) Show that if n = 3k + 2, then 2n³ + 3n² +n=3[(2n + 1)(k+ 1)(n)], which is a multiple of 3.
(c) Find an expression for 2n³ +3n² +n when n = 3k + 1 that shows the polynomial is a multiple of 3.
Together these facts are enough to show (proof by cases) that 2n³ + 3n² +n is always a multiple of 3.
(d) Prove that if an integer m is a multiple of 3 and also an even number, then it must be a multiple of 6.
Note: When we say a is a multiple of b, we are asserting the claim 3k(a = kb). You need to show the
implication [(k(m = 3k)) ^ (³k(m = 2k))] → [³k(m = 6k)]. How are the k's related?
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
