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Consider the statement that 3 divides n³ + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that Basis step: P(1) states that Inductive step: Assume that Show that We have completed the basis step and the inductive step. By mathematical induction, we know that k>0, (P(K)→ P(k+ 1)) is true, that is, vk>0, (3 divides K+2k3 divides (k+ 1)³ + 2(k+ 1)). 3 divided K + 2k for an arbitrary integer k> 0. P(n) is true for all integers n ≥ 1. 3 divides n³ + 2n. 3 divides 1³ +2 1, which is true since 1³ +21=3, and 3 divides 3. Reset Click and drag statements to fill in the details of showing that K(P(K)→ P(k+ 1)) is true, thereby completing the induction step. By the inductive hypothesis, 3 divides (k³ + 5k). 3 divides 3(K + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (K³ + 2k) + 3(k² + k + 1). This completes the inductive step. (k+1)³ + 2(k+ 1) = (k³ + 3k² + 3k + 1) +(2k + 2) (K³ + 5k) + 3(k² + 1) (k+1)³ + 2(k+ 1) = (kª + 3k² + 3k+1)+(2k + 2) (K³+2k) + 3(k² + k + 1) = = By the inductive hypothesis, 3 divides (K³ + 2k). 3 divides 3(k² + k + 1) because it is 3 times an integer.

Consider the statement that 3 divides n³ + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that Basis step: P(1) states that Inductive step: Assume that Show that We have completed the basis step and the inductive step. By mathematical induction, we know that k>0, (P(K)→ P(k+ 1)) is true, that is, vk>0, (3 divides K+2k3 divides (k+ 1)³ + 2(k+ 1)). 3 divided K + 2k for an arbitrary integer k> 0. P(n) is true for all integers n ≥ 1. 3 divides n³ + 2n. 3 divides 1³ +2 1, which is true since 1³ +21=3, and 3 divides 3. Reset Click and drag statements to fill in the details of showing that K(P(K)→ P(k+ 1)) is true, thereby completing the induction step. By the inductive hypothesis, 3 divides (k³ + 5k). 3 divides 3(K + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (K³ + 2k) + 3(k² + k + 1). This completes the inductive step. (k+1)³ + 2(k+ 1) = (k³ + 3k² + 3k + 1) +(2k + 2) (K³ + 5k) + 3(k² + 1) (k+1)³ + 2(k+ 1) = (kª + 3k² + 3k+1)+(2k + 2) (K³+2k) + 3(k² + k + 1) = = By the inductive hypothesis, 3 divides (K³ + 2k). 3 divides 3(k² + k + 1) because it is 3 times an integer.

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Consider the statement that 3 divides n³ + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that Basis step: P(1) states that Inductive step: Assume that Show that We have completed the basis step and the inductive step. By mathematical induction, we know that vk>0, (P(K)→ P(k+ 1)) is true, that is, vk>0, (3 divides k³ + 2k→ 3 divides (k+ 1)³ + 2(k+ 1)). 3 divided K³ + 2k for an arbitrary integer k> 0. P(n) is true for all integers n ≥ 1. 3 divides 1³ +2 3 divides n³ + 2n. 1, which is true since 1³ +21=3, and 3 divides 3. Reset Click and drag statements to fill in the details of showing that VK(P(K) → P(k+ 1)) is true, thereby completing the induction step. By the inductive hypothesis, 3 divides (k³ + 5k). 3 divides 3(K³ + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (k³ + 2k) + 3(k² + k + 1). This completes the inductive step. (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) (k³ + 5k) + 3(K² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) (K³ + 2k) + 3(k² + k + 1) = By the inductive hypothesis, 3 divides (k³ + 2k). 3 divides 3(K² + k + 1) because it is times an integer.
Transcribed Image Text:Consider the statement that 3 divides n³ + 2n whenever n is a positive integer. Outline the proof by clicking and dragging to complete each statement. Let P(n) be the proposition that Basis step: P(1) states that Inductive step: Assume that Show that We have completed the basis step and the inductive step. By mathematical induction, we know that vk>0, (P(K)→ P(k+ 1)) is true, that is, vk>0, (3 divides k³ + 2k→ 3 divides (k+ 1)³ + 2(k+ 1)). 3 divided K³ + 2k for an arbitrary integer k> 0. P(n) is true for all integers n ≥ 1. 3 divides 1³ +2 3 divides n³ + 2n. 1, which is true since 1³ +21=3, and 3 divides 3. Reset Click and drag statements to fill in the details of showing that VK(P(K) → P(k+ 1)) is true, thereby completing the induction step. By the inductive hypothesis, 3 divides (k³ + 5k). 3 divides 3(K³ + 1) because it is 3 times an integer. By part (i) of Theorem 1 in Section 4.1, 3 divides the sum (k³ + 2k) + 3(k² + k + 1). This completes the inductive step. (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) (k³ + 5k) + 3(K² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) (K³ + 2k) + 3(k² + k + 1) = By the inductive hypothesis, 3 divides (k³ + 2k). 3 divides 3(K² + k + 1) because it is times an integer.
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