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Arrange the given steps in the correct order to prove that 3" 6, 3k E search e

Arrange the given steps in the correct order to prove that 3" 6, 3k E search e

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter1: Fundamental Concepts Of Algebra
Section1.2: Exponents And Radicals
Problem 91E
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Please help me with this two questions. I am having trouble understanding why the answers are incorrect and what the correct answer is. Please provide the correct answer to the following 2 questions below

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Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction. Rank the options below. Suppose that for some k> 6, 3k <k! 3k+1 = 3.3k 3k+1 < (k+1).k! For n=7, 37=2187 < 7! = 5040. 3k+1 < (k+1)! 3k+1 < (k+1).3k to search X 4 X 2 6 3 X 5 X < Prev 14 of 16 Next > E
Transcribed Image Text:Arrange the given steps in the correct order to prove that 3" <n! if n is an integer greater than 6, using mathematical induction. Rank the options below. Suppose that for some k> 6, 3k <k! 3k+1 = 3.3k 3k+1 < (k+1).k! For n=7, 37=2187 < 7! = 5040. 3k+1 < (k+1)! 3k+1 < (k+1).3k to search X 4 X 2 6 3 X 5 X < Prev 14 of 16 Next > E
Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: 31 (13+2 1), i.e., 3 | 3, so the basis step is true. INDUCTIVE STEP: Suppose that 3 | (k³ + 2k). By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). 31 (03+2 0), i.e., 3 | 0, so the basis step is true. (k+1)3 + 2(k + 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) (k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3 | (k³ + 2k). By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). (k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) = (k3 + 2k)+3(k² + k + 1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 31 (13+2 1), i.e., 3 | 3, so the basis step is true.
Transcribed Image Text:Click and drag the steps in the correct order to show that 3 divides n³ + 2n whenever n is a positive integer using mathematical induction. BASIS STEP: 31 (13+2 1), i.e., 3 | 3, so the basis step is true. INDUCTIVE STEP: Suppose that 3 | (k³ + 2k). By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). 31 (03+2 0), i.e., 3 | 0, so the basis step is true. (k+1)3 + 2(k + 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) (k+1)3 + 2(k+1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3 | (k³ + 2k). By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). As the sum of two multiples of 3 is again divisible by 3, 3 | ((k+1)3 + 2(k + 1)). (k+1)3 + 2(k+1) = (k³ + 3k² + 3k+ 1) + (2k + 2) = (k3 + 2k)+3(k² + k + 1) By the inductive hypothesis, 3 | (k³ + 2k), and certainly 3 | 3(k² + k + 1). 31 (13+2 1), i.e., 3 | 3, so the basis step is true.
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