n(n?+5) is divisible by 6, for each each integer n20 Proof (by induction) P (0) = " o (0²+5) is divisible by 6" ? 6/0(0?45) = 0 sine 6.0 = 0 / %3D "(K) → PCK+1) : Let k ez Ə K20. Assume P(K) is true that is, " k (k? +5) is divisible by 6". HTS: Kti ((K+1)* +5) is divisible by 6 "K by 6" + 6| K(x²+5) =D3d EZ ə K (K²+5) is divisible 6d= K(x?+5) by defintion of divisibility.
n(n?+5) is divisible by 6, for each each integer n20 Proof (by induction) P (0) = " o (0²+5) is divisible by 6" ? 6/0(0?45) = 0 sine 6.0 = 0 / %3D "(K) → PCK+1) : Let k ez Ə K20. Assume P(K) is true that is, " k (k? +5) is divisible by 6". HTS: Kti ((K+1)* +5) is divisible by 6 "K by 6" + 6| K(x²+5) =D3d EZ ə K (K²+5) is divisible 6d= K(x?+5) by defintion of divisibility.
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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![**Theorem**: \( n(n^2 + 5) \) is divisible by 6, for each integer \( n \geq 0 \).
**Proof (by induction)**:
1. **Base Case** (\( P(0) \)):
\[
P(0) = "0(0^2 + 5) \text{ is divisible by 6}"
\]
Check: \( \frac{6}{0(0^2 + 5)} = 0 \) since \( 6 \cdot 0 = 0 \). Therefore, the base case holds. ✓
2. **Inductive Step** \( P(k) \rightarrow P(k+1) \):
Let \( k \in \mathbb{Z} \) where \( k \geq 0 \). Assume \( P(k) \) is true, that is,
\[
"k(k^2 + 5) \text{ is divisible by 6}".
\]
**Need to Show (NTS)**: \( (k+1)((k+1)^2 + 5) \) is divisible by 6.
Since \( k(k^2 + 5) \) is divisible by 6, it implies:
\[
6 \mid k(k^2 + 5) \Rightarrow \exists d \in \mathbb{Z} \, \exists \, 6d = k(k^2 + 5)
\]
This follows by the definition of divisibility.
This proof effectively employs mathematical induction to demonstrate that the expression \( n(n^2 + 5) \) is divisible by 6 for all integers \( n \geq 0 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdad58c83-f2ea-4146-b813-eac661fb8c00%2F9cd4b7f2-be08-43d7-b894-71ee7dfd96a5%2Ftuscask_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Theorem**: \( n(n^2 + 5) \) is divisible by 6, for each integer \( n \geq 0 \).
**Proof (by induction)**:
1. **Base Case** (\( P(0) \)):
\[
P(0) = "0(0^2 + 5) \text{ is divisible by 6}"
\]
Check: \( \frac{6}{0(0^2 + 5)} = 0 \) since \( 6 \cdot 0 = 0 \). Therefore, the base case holds. ✓
2. **Inductive Step** \( P(k) \rightarrow P(k+1) \):
Let \( k \in \mathbb{Z} \) where \( k \geq 0 \). Assume \( P(k) \) is true, that is,
\[
"k(k^2 + 5) \text{ is divisible by 6}".
\]
**Need to Show (NTS)**: \( (k+1)((k+1)^2 + 5) \) is divisible by 6.
Since \( k(k^2 + 5) \) is divisible by 6, it implies:
\[
6 \mid k(k^2 + 5) \Rightarrow \exists d \in \mathbb{Z} \, \exists \, 6d = k(k^2 + 5)
\]
This follows by the definition of divisibility.
This proof effectively employs mathematical induction to demonstrate that the expression \( n(n^2 + 5) \) is divisible by 6 for all integers \( n \geq 0 \).
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