Problem 8 Use Mathematical Induction to prove that of Clam 1²+ 3² +52 +...+(2n + 1)² = (n + 1)(2n + 1) (2n + 3) 3 whenever n is a nonnegative integer.

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Chapter2: Second-order Linear Odes
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### Problem 8: Mathematical Induction

#### Objective:
Use **Mathematical Induction** to prove the following statement:

\[
1^2 + 3^2 + 5^2 + \cdots + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3}
\]

whenever \( n \) is a nonnegative integer.

#### Explanation
To begin utilizing mathematical induction, we'll typically follow these steps:

1. **Base Case**: Validate the equation for the initial value of \( n \), generally \( n = 0 \).

2. **Inductive Step**: Assume the equation holds for some arbitrary nonnegative integer \( k \). This is the **inductive hypothesis**.

3. **Prove for \( k + 1 \)**: Utilize the inductive hypothesis to demonstrate that if the statement holds for \( n = k \), then it also holds for \( n = k + 1 \).

#### Details of each step will be provided below:

1. **Base Case**:
   - For \( n = 0 \),
   \[
   1^2 = \frac{(0 + 1)(2 \cdot 0 + 1)(2 \cdot 0 + 3)}{3}
   \]
   simplifies to
   \[
   1 = \frac{(1)(1)(3)}{3} = 1
   \]
   Hence, the base case holds true.

2. **Inductive Hypothesis**:
   Assume the equation holds for \( n = k \), i.e.,
   \[
   1^2 + 3^2 + 5^2 + \cdots + (2k + 1)^2 = \frac{(k + 1)(2k + 1)(2k + 3)}{3}
   \]

3. **Inductive Step**:
   Prove the equation for \( n = k + 1 \).

   This means showing:
   \[
   1^2 + 3^2 + 5^2 + \cdots + (2k + 1)^2 + (2(k+1) + 1)^2 = \frac{(k + 2)(2(k + 1) + 1)(2(k
Transcribed Image Text:### Problem 8: Mathematical Induction #### Objective: Use **Mathematical Induction** to prove the following statement: \[ 1^2 + 3^2 + 5^2 + \cdots + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3} \] whenever \( n \) is a nonnegative integer. #### Explanation To begin utilizing mathematical induction, we'll typically follow these steps: 1. **Base Case**: Validate the equation for the initial value of \( n \), generally \( n = 0 \). 2. **Inductive Step**: Assume the equation holds for some arbitrary nonnegative integer \( k \). This is the **inductive hypothesis**. 3. **Prove for \( k + 1 \)**: Utilize the inductive hypothesis to demonstrate that if the statement holds for \( n = k \), then it also holds for \( n = k + 1 \). #### Details of each step will be provided below: 1. **Base Case**: - For \( n = 0 \), \[ 1^2 = \frac{(0 + 1)(2 \cdot 0 + 1)(2 \cdot 0 + 3)}{3} \] simplifies to \[ 1 = \frac{(1)(1)(3)}{3} = 1 \] Hence, the base case holds true. 2. **Inductive Hypothesis**: Assume the equation holds for \( n = k \), i.e., \[ 1^2 + 3^2 + 5^2 + \cdots + (2k + 1)^2 = \frac{(k + 1)(2k + 1)(2k + 3)}{3} \] 3. **Inductive Step**: Prove the equation for \( n = k + 1 \). This means showing: \[ 1^2 + 3^2 + 5^2 + \cdots + (2k + 1)^2 + (2(k+1) + 1)^2 = \frac{(k + 2)(2(k + 1) + 1)(2(k
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