Prove by Mathematical Induction step by step: "Vn € Z+, 1!-3!-5! (2n − 1)! ≥ (n!)*.** -
Prove by Mathematical Induction step by step: "Vn € Z+, 1!-3!-5! (2n − 1)! ≥ (n!)*.** -
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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Discrete Math
![### Proving Factorial Inequality via Mathematical Induction
#### Problem Statement:
Prove by Mathematical Induction step by step:
\[ \forall n \in \mathbb{Z}^+, \, 1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2n - 1)! \geq (n!)^n. \]
#### Solution:
To prove the given inequality using mathematical induction, we follow these steps:
1. **Base Case:**
Validate the statement for the initial value of \( n \).
2. **Inductive Hypothesis:**
Assume the statement is true for some arbitrary positive integer \( k \).
3. **Inductive Step:**
Show that if the statement holds for \( n = k \), then it must also hold for \( n = k + 1 \).
---
#### Detailed Induction Steps:
1. **Base Case:**
Check \( n = 1 \):
\[
\text{Left-hand side (LHS)} = 1!
\]
\[
\text{Right-hand side (RHS)} = (1!)^1 = 1
\]
Clearly, \( 1 \geq 1 \) holds true.
2. **Inductive Hypothesis:**
Assume the inequality holds for \( n = k \):
\[
1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2k - 1)! \geq (k!)^k.
\]
3. **Inductive Step:**
We need to show the inequality holds for \( n = k + 1 \):
\[
1! \cdot 3! \cdot 5! \cdots \cdot (2k - 1)! \cdot (2(k+1) - 1)! \geq ((k+1)!)^{k+1}.
\]
From the inductive hypothesis:
\[
1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2k - 1)! \geq (k!)^k,
\]
Multiply both sides by \((2(k+1) - 1)! = (2k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd7e504e0-f273-40ab-8020-eddc4709afa9%2F3f0d7620-f8b7-4ff7-80a7-52db017e3c28%2Fzqu428_processed.png&w=3840&q=75)
Transcribed Image Text:### Proving Factorial Inequality via Mathematical Induction
#### Problem Statement:
Prove by Mathematical Induction step by step:
\[ \forall n \in \mathbb{Z}^+, \, 1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2n - 1)! \geq (n!)^n. \]
#### Solution:
To prove the given inequality using mathematical induction, we follow these steps:
1. **Base Case:**
Validate the statement for the initial value of \( n \).
2. **Inductive Hypothesis:**
Assume the statement is true for some arbitrary positive integer \( k \).
3. **Inductive Step:**
Show that if the statement holds for \( n = k \), then it must also hold for \( n = k + 1 \).
---
#### Detailed Induction Steps:
1. **Base Case:**
Check \( n = 1 \):
\[
\text{Left-hand side (LHS)} = 1!
\]
\[
\text{Right-hand side (RHS)} = (1!)^1 = 1
\]
Clearly, \( 1 \geq 1 \) holds true.
2. **Inductive Hypothesis:**
Assume the inequality holds for \( n = k \):
\[
1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2k - 1)! \geq (k!)^k.
\]
3. **Inductive Step:**
We need to show the inequality holds for \( n = k + 1 \):
\[
1! \cdot 3! \cdot 5! \cdots \cdot (2k - 1)! \cdot (2(k+1) - 1)! \geq ((k+1)!)^{k+1}.
\]
From the inductive hypothesis:
\[
1! \cdot 3! \cdot 5! \cdot \cdots \cdot (2k - 1)! \geq (k!)^k,
\]
Multiply both sides by \((2(k+1) - 1)! = (2k
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