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
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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 \(n\).
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.
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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