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15) n(n*+5) is divisible by for each integer

15) n(n*+5) is divisible by for each integer

Advanced Engineering Mathematics
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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Are there a way where I can turn 6d = k(k^2 + 5) to 6 ( an integer ) = k+1 ((k+1)^2 + 5)?
The following is a proof by induction showing that \( n(n^2 + 5) \) is divisible by 6 for each integer \( n \geq 0 \). **Proof (by induction)** 1. **Base Case: \( n = 0 \)** \( P(0): 0(0^2 + 5) \) is divisible by 6? \( 6 \cdot 0 = 0 \) since \( 6 \cdot 0 = 0 \). Therefore, true for \( n = 0 \). 2. **Inductive Step: Assume \( P(k) \) is true implies \( P(k+1) \) is true** Let \( k \in \mathbb{Z} \) such that \( k \geq 0 \). Assume \( P(k) \) is true; that is, \( k(k^2 + 5) \) is divisible by 6. So this means that \( 6d = k(k^2 + 5) \) for some integer \( d \), by the definition of divisibility. **Need to Show (NTS):** \( k+1((k+1)^2 + 5) \) is divisible by 6 \( 6 \mid k+1((k+1)^2 + 5) \Rightarrow 6d = k+1((k+1)^2 + 5)\) **Now,** \( 6d = k(k^2 + 5) \) implies that... (followed by further calculations not shown). The proof involves checking the initial base case and then assuming the property holds for \( n = k \) and proving it holds for \( n = k+1 \). Additional steps are needed to complete the inductive step.
Transcribed Image Text:The following is a proof by induction showing that \( n(n^2 + 5) \) is divisible by 6 for each integer \( n \geq 0 \). **Proof (by induction)** 1. **Base Case: \( n = 0 \)** \( P(0): 0(0^2 + 5) \) is divisible by 6? \( 6 \cdot 0 = 0 \) since \( 6 \cdot 0 = 0 \). Therefore, true for \( n = 0 \). 2. **Inductive Step: Assume \( P(k) \) is true implies \( P(k+1) \) is true** Let \( k \in \mathbb{Z} \) such that \( k \geq 0 \). Assume \( P(k) \) is true; that is, \( k(k^2 + 5) \) is divisible by 6. So this means that \( 6d = k(k^2 + 5) \) for some integer \( d \), by the definition of divisibility. **Need to Show (NTS):** \( k+1((k+1)^2 + 5) \) is divisible by 6 \( 6 \mid k+1((k+1)^2 + 5) \Rightarrow 6d = k+1((k+1)^2 + 5)\) **Now,** \( 6d = k(k^2 + 5) \) implies that... (followed by further calculations not shown). The proof involves checking the initial base case and then assuming the property holds for \( n = k \) and proving it holds for \( n = k+1 \). Additional steps are needed to complete the inductive step.
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