Click and drag the steps in the correct order to show that 6 divides n³- n whenever n is a nonnegative integer using mathematicalinduction.BASIS STEP:INDUCTIVE STEP:6 (131), i.e., 610, the basis step is true.61 (03-0), i.e., 6 1 0, so the basis step is true.Suppose that 6 | k³ - k.By the inductive hypothesis, 61 (k³k). Clearly, 313k(k+1), but 3k(k+ 1) is also even as one of kork+ 1 is even; therefore, 6 | 3k(k+ 1) as well.By the inductive hypothesis, 3 I (K³-k). Clearly, 313k(k+1), but 3k(k+ 1) is also odd as one of k or k+ 1 is even; therefore, 6 | 3k(k+ 1) as well.(k+ 1)³(k+ 1) = (k³ + 3k² + 3k + 1) − (k+ 1) =(k³ k) + 3k(k+1)As the sum of two multiples of 6 is again divisibleby 6, 61 ((k+ 1)³ − (k+ 1)).

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Click and drag the steps in the correct order to show that 6 divides n³. - n whenever n is a nonnegative integer using mathematical induction. BASIS STEP: INDUCTIVE STEP: 61 (131), i.e., 610, the basis step is true. 61 (03-0), i.e., 610, so the basis step is true. Suppose that 61 k³ - k. By the inductive hypothesis, 6 I (K³-k). Clearly, 31 3k(k+ 1), but 3k(k+ 1) is also even as one of k or k+ 1 is even; therefore, 6 1 3k(k+ 1) as well. By the inductive hypothesis, 3 I (k³ - k). Clearly, 31 3k(k+ 1), but 3k(k+ 1) is also odd as one of kor k + 1 is even; therefore, 6 | 3k(k+ 1) as well. (k+ 1)³(k+ 1) = (k³ + 3k² + 3k + 1) - (k+ 1) = (ksk) + 3k(k+1) As the sum of two multiples of 6 is again divisible by 6, 61 ((k+ 1)³-(k+ 1)).

Click and drag the steps in the correct order to show that 6 divides n³. - n whenever n is a nonnegative integer using mathematical induction. BASIS STEP: INDUCTIVE STEP: 61 (131), i.e., 610, the basis step is true. 61 (03-0), i.e., 610, so the basis step is true. Suppose that 61 k³ - k. By the inductive hypothesis, 6 I (K³-k). Clearly, 31 3k(k+ 1), but 3k(k+ 1) is also even as one of k or k+ 1 is even; therefore, 6 1 3k(k+ 1) as well. By the inductive hypothesis, 3 I (k³ - k). Clearly, 31 3k(k+ 1), but 3k(k+ 1) is also odd as one of kor k + 1 is even; therefore, 6 | 3k(k+ 1) as well. (k+ 1)³(k+ 1) = (k³ + 3k² + 3k + 1) - (k+ 1) = (ksk) + 3k(k+1) As the sum of two multiples of 6 is again divisible by 6, 61 ((k+ 1)³-(k+ 1)).

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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Question

Click and drag the steps in the correct order to show that 6 divides n³- n whenever n is a nonnegative integer using mathematical
induction.
BASIS STEP:
INDUCTIVE STEP:
6 (131), i.e., 610, the basis step is true.
61 (03-0), i.e., 6 1 0, so the basis step is true.
Suppose that 6 | k³ - k.
By the inductive hypothesis, 61 (k³k). Clearly, 31
3k(k+1), but 3k(k+ 1) is also even as one of kor
k+ 1 is even; therefore, 6 | 3k(k+ 1) as well.
By the inductive hypothesis, 3 I (K³-k). Clearly, 31
3k(k+1), but 3k(k+ 1) is also odd as one of k or k
+ 1 is even; therefore, 6 | 3k(k+ 1) as well.
(k+ 1)³(k+ 1) = (k³ + 3k² + 3k + 1) − (k+ 1) =
(k³ k) + 3k(k+1)
As the sum of two multiples of 6 is again divisible
by 6, 61 ((k+ 1)³ − (k+ 1)).

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