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Use induction to prove: for any integer n > 1, 3|(n – 4n + 6). Hint: this question is case sensitive, you should ONLY use k, but not K in your answers. Base case n = ,n? - 4n + 6 = which is divisible by 3. Inductive step Assume that for any k > , 3| , we will prove that 3| By the inductive hypothesis, there exists an integer m such that = 3m. It follows that k3 %3D (k + 1)3 – 4(k + 1) + 6 = k³+ k2+ k+

Use induction to prove: for any integer n > 1, 3|(n – 4n + 6). Hint: this question is case sensitive, you should ONLY use k, but not K in your answers. Base case n = ,n? - 4n + 6 = which is divisible by 3. Inductive step Assume that for any k > , 3| , we will prove that 3| By the inductive hypothesis, there exists an integer m such that = 3m. It follows that k3 %3D (k + 1)3 – 4(k + 1) + 6 = k³+ k2+ k+

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter2: The Integers
Section2.2: Mathematical Induction
Problem 46E: Use generalized induction and Exercise 43 to prove that n22n for all integers n5. (In connection...
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Use induction to prove: for any integer n > 1, 3|(n – 4n + 6). - Hint: this question is case sensitive, you should ONLY use k, but not K in your answers. Base case n = , n? – 4n + 6 = which is divisible by 3. Inductive step Assume that for any k > , 3| , we will prove that 3| By the inductive hypothesis, there exists an integer m such that = 3m. It follows that k3 : (k + 1)3 – 4(k + 1) + 6 = k³+ k²+ k+ By inductive hypothesis = 3( Since m and k are integers, must be an integer. Hence, 3|
Transcribed Image Text:Use induction to prove: for any integer n > 1, 3|(n – 4n + 6). - Hint: this question is case sensitive, you should ONLY use k, but not K in your answers. Base case n = , n? – 4n + 6 = which is divisible by 3. Inductive step Assume that for any k > , 3| , we will prove that 3| By the inductive hypothesis, there exists an integer m such that = 3m. It follows that k3 : (k + 1)3 – 4(k + 1) + 6 = k³+ k²+ k+ By inductive hypothesis = 3( Since m and k are integers, must be an integer. Hence, 3|
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