1. Show that 2! 4! 6! (2n)! ≥ ((n + 1)!)" for ne z+. . ...
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Problem 1RQ
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Please follow exactly the solution steps of the 2nd image to answer my question. The way you answer my question should follow the steps on that image
Transcribed Image Text:**Learning Target R4 Core**: *Given a statement to be proven by (weak) induction, I can state and prove the base case, state the inductive hypothesis, and outline the proof. I can describe the subtle difference between 'weak' and 'strong' induction.*
For each statement below, complete each of the following parts:
- **State and prove the base case**
- **State the inductive hypothesis**
- **Outline how the rest of the proof would go**, that is, tell me roughly what will happen to complete the proof, identifying how this section of the proof will begin and ultimately what it will look like when you're done (as modeled in class videos); you do not need to actually do the complete proof.
You might not be able to prove these claims with what we learned in this class, and that's okay. You're just setting it up!
Remember if you have a question about this, you can email/send me a message on Teams. I want to help you succeed!
1. **Show that** \( 2! \cdot 4! \cdot 6! \cdots (2n)! \geq ((n+1)!)^n \) *for* \( n \in \mathbb{Z}^+ \).
2. **Show that** \( a + (a+d) + (a+2d) + (a+3d) + \cdots + (a+(n-1)d) = \frac{n(2a + (n-1)d)}{2} \) *for* \( n \in \mathbb{Z}^+ \).
![## Mathematical Induction Proof
### (1) Proof by Mathematical Induction
We aim to prove that for all positive integers \( n \), the following identity holds:
\[ 1^3 + 2^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2. \]
#### Base Case (\( n = 1 \)):
- **Left-Hand Side (LHS):** \(1^3 = 1\)
- **Right-Hand Side (RHS):** \(\left( \frac{1(1+1)}{2} \right)^2 = 1^2 = 1\)
The identity holds for \( n = 1 \).
#### Inductive Hypothesis:
Assume the identity holds for some positive integer \( k \), i.e.,
\[ 1^3 + 2^3 + \ldots + k^3 = \left( \frac{k(k+1)}{2} \right)^2. \]
#### Inductive Step:
Prove that the identity holds for \( k + 1 \), i.e.,
\[ 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \left( \frac{(k+1)(k+2)}{2} \right)^2. \]
Start from the assumption and add \((k+1)^3\) to both sides:
\[ 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3. \]
Simplify the RHS:
\[ \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3. \]
To prove the identity, show that the LHS and RHS are equal:
\[ \frac{k^2(k+1)^2}{4} + (k+1)^3 = \left( \frac{(k+1)(k+2)}{2} \right)^2. \]
Now, manipulate the LHS:
\[
\frac{k^2(k+1)^2}{](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd375a7e5-ad67-4b21-afad-3fb7503a0e18%2Fd88af08b-f04b-4252-8029-99fdfa10d785%2Fyezoza_processed.jpeg&w=3840&q=75)
Transcribed Image Text:## Mathematical Induction Proof
### (1) Proof by Mathematical Induction
We aim to prove that for all positive integers \( n \), the following identity holds:
\[ 1^3 + 2^3 + \ldots + n^3 = \left( \frac{n(n+1)}{2} \right)^2. \]
#### Base Case (\( n = 1 \)):
- **Left-Hand Side (LHS):** \(1^3 = 1\)
- **Right-Hand Side (RHS):** \(\left( \frac{1(1+1)}{2} \right)^2 = 1^2 = 1\)
The identity holds for \( n = 1 \).
#### Inductive Hypothesis:
Assume the identity holds for some positive integer \( k \), i.e.,
\[ 1^3 + 2^3 + \ldots + k^3 = \left( \frac{k(k+1)}{2} \right)^2. \]
#### Inductive Step:
Prove that the identity holds for \( k + 1 \), i.e.,
\[ 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \left( \frac{(k+1)(k+2)}{2} \right)^2. \]
Start from the assumption and add \((k+1)^3\) to both sides:
\[ 1^3 + 2^3 + \ldots + k^3 + (k+1)^3 = \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3. \]
Simplify the RHS:
\[ \left( \frac{k(k+1)}{2} \right)^2 + (k+1)^3 = \frac{k^2(k+1)^2}{4} + (k+1)^3. \]
To prove the identity, show that the LHS and RHS are equal:
\[ \frac{k^2(k+1)^2}{4} + (k+1)^3 = \left( \frac{(k+1)(k+2)}{2} \right)^2. \]
Now, manipulate the LHS:
\[
\frac{k^2(k+1)^2}{
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