Suppose that n is a positive integer. Arrange the steps in the correct order to show that 2 divides n²+ n using mathematical induction. Rank the options below. As the sum of two multiples of 2 is again a multiple of 2, 2 divides (k+ 1)2 + (k+1). (k+ 1)2 + (k+1)= (k² + 2k + 1) + (k+ 1) = (k² + k) + 2(k+1) The inductive hypothesis is that 2 divides k² + k. 12+1=2 and 212, so the basis step is clear. 2 divides K² + k by inductive hypothesis, and clearly 2 divides 2(k+ 1). V
Suppose that n is a positive integer. Arrange the steps in the correct order to show that 2 divides n²+ n using mathematical induction. Rank the options below. As the sum of two multiples of 2 is again a multiple of 2, 2 divides (k+ 1)2 + (k+1). (k+ 1)2 + (k+1)= (k² + 2k + 1) + (k+ 1) = (k² + k) + 2(k+1) The inductive hypothesis is that 2 divides k² + k. 12+1=2 and 212, so the basis step is clear. 2 divides K² + k by inductive hypothesis, and clearly 2 divides 2(k+ 1). V
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter7: Exponents And Exponential Functions
Section7.10: Recursive Formulas
Problem 37PFA
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Transcribed Image Text:Suppose that n is a positive integer.
Arrange the steps in the correct order to show that 2 divides n² + n using mathematical induction.
Rank the options below.
As the sum of two multiples of 2 is again a multiple of 2, 2 divides (k+ 1)² + (k+ 1).
(k+ 1)² + (k+ 1) = (k² + 2k + 1) + (k + 1) = (k² + k) + 2(k+1)
The inductive hypothesis is that 2 divides k² + k.
12+1=2 and 212, so the basis step is clear.
2 divides k² + k by inductive hypothesis, and clearly 2 divides 2(k + 1).
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