5. For all positive integer n, n³ +5n+6 is divis We use induction on n Then (n+12²³+ 5(n+1) 76 = n²³² +31²³²2w 135 (n+1) (ht? 1²72+1 (1)
5. For all positive integer n, n³ +5n+6 is divis We use induction on n Then (n+12²³+ 5(n+1) 76 = n²³² +31²³²2w 135 (n+1) (ht? 1²72+1 (1)
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
Use induction method
![**Mathematical Induction Problem:**
5. **Problem Statement**
- For all positive integers \( n \), prove that \( n^3 + 5n + 6 \) is divisible by 3.
**Solution Approach:**
- **Induction Hypothesis:** Assume for an integer \( n \), the expression \( n^3 + 5n + 6 \) is divisible by 3.
- **Basis Step:** Verify for \( n = 1 \):
\[
1^3 + 5 \times 1 + 6 = 1 + 5 + 6 = 12
\]
12 is divisible by 3.
- **Inductive Step:** Assume true for \( n = k \), i.e., \( k^3 + 5k + 6 \) is divisible by 3.
- Show that it holds for \( n = k + 1 \):
\[
(k+1)^3 + 5(k+1) + 6 = k^3 + 3k^2 + 3k + 1 + 5k + 5 + 6
\]
\[
= k^3 + 5k + 6 + 3k^2 + 3k + 12
\]
The expression can be rearranged to show divisibility by 3.
Thus, by induction, \( n^3 + 5n + 6 \) is divisible by 3 for all positive integers \( n \).
**Conclusion:**
The problem demonstrates a typical use of mathematical induction to prove divisibility properties for polynomial expressions dependent on integer variables.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad4ca8bb-6fc2-4881-9132-73064896bc20%2F49f62616-5d12-4df8-90bc-2a4ffbabbfc2%2Fcvxcb64_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Mathematical Induction Problem:**
5. **Problem Statement**
- For all positive integers \( n \), prove that \( n^3 + 5n + 6 \) is divisible by 3.
**Solution Approach:**
- **Induction Hypothesis:** Assume for an integer \( n \), the expression \( n^3 + 5n + 6 \) is divisible by 3.
- **Basis Step:** Verify for \( n = 1 \):
\[
1^3 + 5 \times 1 + 6 = 1 + 5 + 6 = 12
\]
12 is divisible by 3.
- **Inductive Step:** Assume true for \( n = k \), i.e., \( k^3 + 5k + 6 \) is divisible by 3.
- Show that it holds for \( n = k + 1 \):
\[
(k+1)^3 + 5(k+1) + 6 = k^3 + 3k^2 + 3k + 1 + 5k + 5 + 6
\]
\[
= k^3 + 5k + 6 + 3k^2 + 3k + 12
\]
The expression can be rearranged to show divisibility by 3.
Thus, by induction, \( n^3 + 5n + 6 \) is divisible by 3 for all positive integers \( n \).
**Conclusion:**
The problem demonstrates a typical use of mathematical induction to prove divisibility properties for polynomial expressions dependent on integer variables.
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