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.
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.
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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![### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe872339c-719c-4a81-bc7c-37653a99cb41%2Ff8e1e304-b91b-42ed-a560-8f699145b53a%2F9ureryf_processed.jpeg&w=3840&q=75)
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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