Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 1)² 22 +42 +62+...+(2(k+1))² = = ... +(2(k+1))² = 2k(k+ 1)(2k + 1) 3 2(k+1) 2(k+ 1) (k+ Rewrite the right-hand side in the desired form. 22 +42 +62 +. 3 + 12 3 (2k(2k + 1) + 12 3 1) 3 )(2k + So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n. Need Help?
Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 1)² 22 +42 +62+...+(2(k+1))² = = ... +(2(k+1))² = 2k(k+ 1)(2k + 1) 3 2(k+1) 2(k+ 1) (k+ Rewrite the right-hand side in the desired form. 22 +42 +62 +. 3 + 12 3 (2k(2k + 1) + 12 3 1) 3 )(2k + So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n. Need Help?
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Title: Mathematical Induction Example: Sum of Even Squares**
**Objective:** Rewrite the right-hand side as a single fraction, and then factor the numerator completely.
**Step-by-Step Solution:**
Given equation:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \frac{2k(k+1)(2k+1) + 12(\text{ })}{3} \]
1. **Rewrite Equation:**
\[
= \frac{(\text{ })2k(2k+1) + 12(\text{ })}{3}
\]
2. **Factor Out Common Terms:**
\[
= \frac{2(k+1)(\text{ })}{3}
\]
3. **Expand and Rearrange:**
\[
= \frac{2(k+1)(k + \text{ })(2k + \text{ })}{3}
\]
**Final Expression:**
Rewrite the right-hand side in the desired form:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \text{ } \]
**Conclusion:**
So, \(P_{k+1}\) is true. We conclude by the principle of mathematical induction that \(P_n\) is true for all natural numbers \(n\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f36f111-f1bc-4b8c-9508-d8f50a3e95e1%2Fddee6782-0d4e-48bb-a18f-443100ba1973%2Fotyheb7_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title: Mathematical Induction Example: Sum of Even Squares**
**Objective:** Rewrite the right-hand side as a single fraction, and then factor the numerator completely.
**Step-by-Step Solution:**
Given equation:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \frac{2k(k+1)(2k+1) + 12(\text{ })}{3} \]
1. **Rewrite Equation:**
\[
= \frac{(\text{ })2k(2k+1) + 12(\text{ })}{3}
\]
2. **Factor Out Common Terms:**
\[
= \frac{2(k+1)(\text{ })}{3}
\]
3. **Expand and Rearrange:**
\[
= \frac{2(k+1)(k + \text{ })(2k + \text{ })}{3}
\]
**Final Expression:**
Rewrite the right-hand side in the desired form:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \text{ } \]
**Conclusion:**
So, \(P_{k+1}\) is true. We conclude by the principle of mathematical induction that \(P_n\) is true for all natural numbers \(n\).
![**Using Mathematical Induction for Sum of Squares Formula**
**Statement:**
Use the principle of mathematical induction to show that the following statement is true for all natural numbers:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3} \]
**Inductive Step Definitions:**
Let \( P_n \) denote the statement:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3} \]
**Base Case:**
Check that \( P_1 \) is true:
\[ 2^2 = 4 \]
\[ 2\left( \frac{1(1 + 1)(2 \times 1 + 1)}{3} \right) = 4 \]
Thus, \( P_1 \) is true.
**Inductive Hypothesis:**
Assume \( P_k \) is true:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2k)^2 = \frac{2k(k + 1)(2k + 1)}{3} \]
**Inductive Step:**
To show that \( P_{k+1} \) is true, add \( (2(k + 1))^2 \) to both sides of \( P_k \):
\[ 2^2 + 4^2 + 6^2 + \ldots + (2k)^2 + (2(k+1))^2 \]
\[ = \frac{2k(k + 1)(2k + 1)}{3} + (2(k+1))^2 \]
**Simplification:**
Rewrite the right-hand side as a single fraction and factor the numerator completely:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \frac{2k(k + 1)(2k + 1)}{3} + \frac{12(k+1)^2}{3} \]
This establishes the inductive step, and thus the formula is proven for all natural numbers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9f36f111-f1bc-4b8c-9508-d8f50a3e95e1%2Fddee6782-0d4e-48bb-a18f-443100ba1973%2Fvis4tmc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Using Mathematical Induction for Sum of Squares Formula**
**Statement:**
Use the principle of mathematical induction to show that the following statement is true for all natural numbers:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3} \]
**Inductive Step Definitions:**
Let \( P_n \) denote the statement:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2n)^2 = \frac{2n(n + 1)(2n + 1)}{3} \]
**Base Case:**
Check that \( P_1 \) is true:
\[ 2^2 = 4 \]
\[ 2\left( \frac{1(1 + 1)(2 \times 1 + 1)}{3} \right) = 4 \]
Thus, \( P_1 \) is true.
**Inductive Hypothesis:**
Assume \( P_k \) is true:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2k)^2 = \frac{2k(k + 1)(2k + 1)}{3} \]
**Inductive Step:**
To show that \( P_{k+1} \) is true, add \( (2(k + 1))^2 \) to both sides of \( P_k \):
\[ 2^2 + 4^2 + 6^2 + \ldots + (2k)^2 + (2(k+1))^2 \]
\[ = \frac{2k(k + 1)(2k + 1)}{3} + (2(k+1))^2 \]
**Simplification:**
Rewrite the right-hand side as a single fraction and factor the numerator completely:
\[ 2^2 + 4^2 + 6^2 + \ldots + (2(k+1))^2 = \frac{2k(k + 1)(2k + 1)}{3} + \frac{12(k+1)^2}{3} \]
This establishes the inductive step, and thus the formula is proven for all natural numbers.
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