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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd49a5070-9d99-4a2f-a144-be47e7ddead5%2Fd2e5fdca-6b36-44ee-82df-4d6a165baa5a%2Fzaggypg_processed.png&w=3840&q=75)
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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