Problem 3 Use Mathematical Induction to prove that 1² + 3² + 5²+...+ (2n + 1)² = (n + 1)(2n + 1) (2n + 3) 3 whenever n is a nonnegative integer. Solution 2
Problem 3 Use Mathematical Induction to prove that 1² + 3² + 5²+...+ (2n + 1)² = (n + 1)(2n + 1) (2n + 3) 3 whenever n is a nonnegative integer. Solution 2
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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![### Mathematical Induction Proof of Sum of Squares Formula
#### Problem 3
**Use Mathematical Induction to prove that**
\[ 1^2 + 3^2 + 5^2 + \cdots + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3} \]
**whenever \( n \) is a nonnegative integer.**
#### Solution:
\[
\begin{aligned}
& \text{Base case: Let } n = 0 \\
& (2 \cdot 0 + 1)^2 = \frac{(0 + 1)(2 \cdot 0 + 1)(2 \cdot 0 + 3)}{3} \\
& 1^2 = \frac{(1)(1)(3)}{3} \\
& 1 = 1 \text{, which is true.}
\end{aligned}
\]
Next step is to assume the statement is true for some \( k \) and then prove it for \( k+1 \). This will be the inductive step.
\[
\text{Assume } \sum_{i=0}^k (2i + 1)^2 = \frac{(k + 1)(2k + 1)(2k + 3)}{3}
\]
Then we need to show:
\[
\sum_{i=0}^{k+1} (2i + 1)^2 = \frac{(k + 2)(2(k + 1) + 1)(2(k + 1) + 3)}{3}
\]
Add the next term \( (2(k+1) + 1)^2 \) to both sides of the assumed statement and simplify.
Completing the details of the induction step ultimately demonstrates that the formula works for \( k+1 \), and by induction, it holds for all nonnegative integers \( n \).
Thus, we have used mathematical induction to prove the given formula.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbe52e616-7bd7-4d85-a476-7e5fcec53ee5%2F5586c692-10d1-4738-9412-9e4f805fddaa%2Fyw4c453_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Mathematical Induction Proof of Sum of Squares Formula
#### Problem 3
**Use Mathematical Induction to prove that**
\[ 1^2 + 3^2 + 5^2 + \cdots + (2n + 1)^2 = \frac{(n + 1)(2n + 1)(2n + 3)}{3} \]
**whenever \( n \) is a nonnegative integer.**
#### Solution:
\[
\begin{aligned}
& \text{Base case: Let } n = 0 \\
& (2 \cdot 0 + 1)^2 = \frac{(0 + 1)(2 \cdot 0 + 1)(2 \cdot 0 + 3)}{3} \\
& 1^2 = \frac{(1)(1)(3)}{3} \\
& 1 = 1 \text{, which is true.}
\end{aligned}
\]
Next step is to assume the statement is true for some \( k \) and then prove it for \( k+1 \). This will be the inductive step.
\[
\text{Assume } \sum_{i=0}^k (2i + 1)^2 = \frac{(k + 1)(2k + 1)(2k + 3)}{3}
\]
Then we need to show:
\[
\sum_{i=0}^{k+1} (2i + 1)^2 = \frac{(k + 2)(2(k + 1) + 1)(2(k + 1) + 3)}{3}
\]
Add the next term \( (2(k+1) + 1)^2 \) to both sides of the assumed statement and simplify.
Completing the details of the induction step ultimately demonstrates that the formula works for \( k+1 \), and by induction, it holds for all nonnegative integers \( n \).
Thus, we have used mathematical induction to prove the given formula.
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