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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: INDUCTIVE STEP: 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., 310, so the basis step is true. (k+1)+2(k+1)=(k³ + 3k² + 1) + (2k+ 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3|(k³ + 2k). (k+1)+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., 313, so the basis step is true. Reset

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: INDUCTIVE STEP: 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., 310, so the basis step is true. (k+1)+2(k+1)=(k³ + 3k² + 1) + (2k+ 2) = (k³ + 2k) + 3(k² + 1) Suppose that 3|(k³ + 2k). (k+1)+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., 313, so the basis step is true. Reset

Advanced Engineering Mathematics
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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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: 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)). INDUCTIVE STEP: 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) Suppose that 3 | (k³ + 2k). (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). 3|(13+2 1), i.e., 3 | 3, so the basis step is true. Reset
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: 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)). INDUCTIVE STEP: 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) Suppose that 3 | (k³ + 2k). (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). 3|(13+2 1), i.e., 3 | 3, so the basis step is true. Reset
Arrange the steps in the correct order to prove that 21 divides 4n+1+52n-1 whenever n is a positive integer. Rank the options below. 4-4(k+1)+(4+21) - 52k-1 = 4(4(k+1)+52k-1)+21.52k-1 Hence, 211 (4(k+1)+1+52(k+1) - 1), and by mathematical induction, 211 (4+1 + 52n-1) whenever n is a positive integer. As 21141+1+52-1-1=21, the basis step is true. For the inductive hypothesis, suppose that 21 | (4k+1+52k-1). 4.4(k+1)+25.52k-1=4.4(k+1)+(4+21). 52k-1 4(k+1)+1+52(k+1)-1=4.4(k+1)+25.52k-1 The term 4(4(k+1)+52k-1) is divisible by 21 according to the inductive hypothesis, and the term 21.52k-1 is clearly divisible by 21.
Transcribed Image Text:Arrange the steps in the correct order to prove that 21 divides 4n+1+52n-1 whenever n is a positive integer. Rank the options below. 4-4(k+1)+(4+21) - 52k-1 = 4(4(k+1)+52k-1)+21.52k-1 Hence, 211 (4(k+1)+1+52(k+1) - 1), and by mathematical induction, 211 (4+1 + 52n-1) whenever n is a positive integer. As 21141+1+52-1-1=21, the basis step is true. For the inductive hypothesis, suppose that 21 | (4k+1+52k-1). 4.4(k+1)+25.52k-1=4.4(k+1)+(4+21). 52k-1 4(k+1)+1+52(k+1)-1=4.4(k+1)+25.52k-1 The term 4(4(k+1)+52k-1) is divisible by 21 according to the inductive hypothesis, and the term 21.52k-1 is clearly divisible by 21.
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