Problem 12 Use Mathematical Induction to prove that 6 n³. -n whenever n is a positive integer
Problem 12 Use Mathematical Induction to prove that 6 n³. -n whenever n is a positive integer
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
![### Problem 12
**Use Mathematical Induction to prove that \(6 \mid (n^3 - n)\) whenever \(n\) is a positive integer.**
For this problem, let's break it down into simple steps to understand the method of proof by mathematical induction.
1. **Base Case:**
- Verify the initial case, often \(n = 1\).
2. **Inductive Step:**
- Assume that the given statement is true for some arbitrary positive integer \(k\). This is called the induction hypothesis.
- Using this hypothesis, prove that the statement is true for \(k + 1\).
If both of these steps are satisfied, then by the principle of mathematical induction, the statement is true for all positive integers \(n\).
Let's begin the proof process.
#### Base Case:
For \(n = 1\),
\[ 6 \mid (1^3 - 1) \implies 6 \mid 0 \]
Which is obviously true, as 0 is divisible by 6.
#### Inductive Step:
Assume that for some positive integer \(k\),
\[ 6 \mid (k^3 - k) \]
This means there exists some integer \(m\) such that:
\[ k^3 - k = 6m \]
We need to prove that \(6 \mid ((k + 1)^3 - (k + 1))\).
\[ (k + 1)^3 - (k + 1) = k^3 + 3k^2 + 3k + 1 - k - 1 \]
\[ = k^3 + 3k^2 + 2k \]
By the induction hypothesis,
\[ k^3 - k = 6m \]
So,
\[ k^3 = 6m + k \]
Substituting this into our equation,
\[ (k + 1)^3 - (k + 1) = (6m + k) + 3k^2 + 2k \]
\[ = 6m + 3k^2 + 3k = 6m + 3k(k + 1) \]
Notice that \(3k(k + 1)\) is always divisible by 6, since \(k(k + 1)\) is the product of two consecutive integers, one of which is always even (thus divisible by](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe872339c-719c-4a81-bc7c-37653a99cb41%2Fdd8105a0-26e6-410f-846d-1e4ba0067fe6%2Fy4hng8r_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem 12
**Use Mathematical Induction to prove that \(6 \mid (n^3 - n)\) whenever \(n\) is a positive integer.**
For this problem, let's break it down into simple steps to understand the method of proof by mathematical induction.
1. **Base Case:**
- Verify the initial case, often \(n = 1\).
2. **Inductive Step:**
- Assume that the given statement is true for some arbitrary positive integer \(k\). This is called the induction hypothesis.
- Using this hypothesis, prove that the statement is true for \(k + 1\).
If both of these steps are satisfied, then by the principle of mathematical induction, the statement is true for all positive integers \(n\).
Let's begin the proof process.
#### Base Case:
For \(n = 1\),
\[ 6 \mid (1^3 - 1) \implies 6 \mid 0 \]
Which is obviously true, as 0 is divisible by 6.
#### Inductive Step:
Assume that for some positive integer \(k\),
\[ 6 \mid (k^3 - k) \]
This means there exists some integer \(m\) such that:
\[ k^3 - k = 6m \]
We need to prove that \(6 \mid ((k + 1)^3 - (k + 1))\).
\[ (k + 1)^3 - (k + 1) = k^3 + 3k^2 + 3k + 1 - k - 1 \]
\[ = k^3 + 3k^2 + 2k \]
By the induction hypothesis,
\[ k^3 - k = 6m \]
So,
\[ k^3 = 6m + k \]
Substituting this into our equation,
\[ (k + 1)^3 - (k + 1) = (6m + k) + 3k^2 + 2k \]
\[ = 6m + 3k^2 + 3k = 6m + 3k(k + 1) \]
Notice that \(3k(k + 1)\) is always divisible by 6, since \(k(k + 1)\) is the product of two consecutive integers, one of which is always even (thus divisible by
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 3 steps with 11 images
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,
