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 **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 InductionWe 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}{

To prove the inequality for n∈Z+, we can use mathematical induction.Base Case: Let's check the…

Answered: 1. Show that 2! 4! 6! (2n)! ≥ ((n +…

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1. Show that 2! 4! 6! (2n)! ≥ ((n + 1)!)" for ne z+. . ...

Advanced Engineering Mathematics

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ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

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Question

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

**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}{

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