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Use the Generalized Principle of Mathematical Induction (PMI) to prove the following. n! > 2n+2 for all n ≥ 6

Use the Generalized Principle of Mathematical Induction (PMI) to prove the following. n! > 2n+2 for all n ≥ 6

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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### Proving Inequalities using the Generalized Principle of Mathematical Induction #### Problem Statement Use the Generalized Principle of Mathematical Induction (PMI) to prove the following inequality: \[ n! > 2^{n+2} \quad \text{for all} \quad n \geq 6 \] #### Solution Outline To prove this inequality, we will follow the steps of mathematical induction: 1. **Base Case:** Verify that the inequality holds for the smallest value of \( n \). 2. **Inductive Step:** Assume the inequality holds for some \( n = k \) and use this assumption to prove that it also holds for \( n = k + 1 \). Let's proceed with the proof: --- ### Step 1: Base Case Check the base case \( n = 6 \). Calculate both sides of the inequality: \[ 6! = 720 \] \[ 2^{6 + 2} = 2^8 = 256 \] Clearly, \[ 720 > 256 \] Therefore, the base case holds. ### Step 2: Inductive Step Assume that the inequality is true for some \( n = k \), i.e., \[ k! > 2^{k+2} \] We need to show that the inequality is also true for \( n = k + 1 \), i.e., \[ (k+1)! > 2^{(k+1) + 2} \] Start from the inductive hypothesis: \[ (k+1)! = (k+1) \cdot k! \] Using the inductive hypothesis \( k! > 2^{k+2} \), we get: \[ (k+1)! > (k+1) \cdot 2^{k+2} \] Now, we need to prove: \[ (k+1) \cdot 2^{k+2} > 2^{(k+1) + 2} = 2^{k+3} \] This simplifies to: \[ (k+1) > 2 \] Clearly, for \( k \geq 6 \), \[ (k+1) \geq 7 \] which is always greater than 2. Therefore, the inequality holds for \( n = k + 1 \). ### Conclusion By the principle of mathematical induction, the inequality \( n
Transcribed Image Text:### Proving Inequalities using the Generalized Principle of Mathematical Induction #### Problem Statement Use the Generalized Principle of Mathematical Induction (PMI) to prove the following inequality: \[ n! > 2^{n+2} \quad \text{for all} \quad n \geq 6 \] #### Solution Outline To prove this inequality, we will follow the steps of mathematical induction: 1. **Base Case:** Verify that the inequality holds for the smallest value of \( n \). 2. **Inductive Step:** Assume the inequality holds for some \( n = k \) and use this assumption to prove that it also holds for \( n = k + 1 \). Let's proceed with the proof: --- ### Step 1: Base Case Check the base case \( n = 6 \). Calculate both sides of the inequality: \[ 6! = 720 \] \[ 2^{6 + 2} = 2^8 = 256 \] Clearly, \[ 720 > 256 \] Therefore, the base case holds. ### Step 2: Inductive Step Assume that the inequality is true for some \( n = k \), i.e., \[ k! > 2^{k+2} \] We need to show that the inequality is also true for \( n = k + 1 \), i.e., \[ (k+1)! > 2^{(k+1) + 2} \] Start from the inductive hypothesis: \[ (k+1)! = (k+1) \cdot k! \] Using the inductive hypothesis \( k! > 2^{k+2} \), we get: \[ (k+1)! > (k+1) \cdot 2^{k+2} \] Now, we need to prove: \[ (k+1) \cdot 2^{k+2} > 2^{(k+1) + 2} = 2^{k+3} \] This simplifies to: \[ (k+1) > 2 \] Clearly, for \( k \geq 6 \), \[ (k+1) \geq 7 \] which is always greater than 2. Therefore, the inequality holds for \( n = k + 1 \). ### Conclusion By the principle of mathematical induction, the inequality \( n
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