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Click and drag the steps in the correct order to show that 3 divides nº +2n whenever n is a positive integer using math induction. BASIS STEP: INDUCTIVE STEP: By the inductive hypothesis, 31 (k³ + 2k), and certainly 3 1 3(k² + 1). Suppose that 31 (k³ + 2k). By the inductive hypothesis, 31 (K³ + 2k), and certainly 3 1 3(k² + k + 1). 31 (13 +21), i.e., 313, so the basis step is true. As the sum of two multiples of 3 is again divisible by 3, 31 ((k+ 1)³ + 2(k+ 1)). (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) = (k³ + 2k) + 3(k² + k + 1) 31 (0³+20), i.e., 3 1 0, so the basis step is true.

Click and drag the steps in the correct order to show that 3 divides nº +2n whenever n is a positive integer using math induction. BASIS STEP: INDUCTIVE STEP: By the inductive hypothesis, 31 (k³ + 2k), and certainly 3 1 3(k² + 1). Suppose that 31 (k³ + 2k). By the inductive hypothesis, 31 (K³ + 2k), and certainly 3 1 3(k² + k + 1). 31 (13 +21), i.e., 313, so the basis step is true. As the sum of two multiples of 3 is again divisible by 3, 31 ((k+ 1)³ + 2(k+ 1)). (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(k² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k+1)+(2k + 2) = (k³ + 2k) + 3(k² + k + 1) 31 (0³+20), i.e., 3 1 0, so the basis step is true.

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: INDUCTIVE STEP: By the inductive hypothesis, 3 I (k³ + 2k), and certainly 3 I 3(k² + 1). Suppose that 31 (k³ + 2k). By the inductive hypothesis, 31 (k³ + 2k), and certainly 3 | 3(k² + k + 1). 31 (13 +21), i.e., 3 1 3, so the basis step is true. As the sum of two multiples of 3 is again divisible by 3, 31 ((k+ 1)³ + 2(k+ 1)). (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(K² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k + 1) + (2k + 2) = (k³ + 2k) + 3(k² + k + 1) 31 (0³+20), i.e., 3 1 0, 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: INDUCTIVE STEP: By the inductive hypothesis, 3 I (k³ + 2k), and certainly 3 I 3(k² + 1). Suppose that 31 (k³ + 2k). By the inductive hypothesis, 31 (k³ + 2k), and certainly 3 | 3(k² + k + 1). 31 (13 +21), i.e., 3 1 3, so the basis step is true. As the sum of two multiples of 3 is again divisible by 3, 31 ((k+ 1)³ + 2(k+ 1)). (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 1) + (2k + 2) = (k³ + 2k) + 3(K² + 1) (k+ 1)³ + 2(k+ 1) = (k³ + 3k² + 3k + 1) + (2k + 2) = (k³ + 2k) + 3(k² + k + 1) 31 (0³+20), i.e., 3 1 0, so the basis step is true.
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