Use induction to prove: for any integer n ≥ 1,(6-4)= 3n² - n. Base case n = Inductive step Assume that for any k > k+1 Σ(6-4)= (6j - 4)+ j=1 j=1 = Σ(6; – 4) = j=1 = =( = 3 k² + k² + k+ k+ j=1 j=1 ,3n²-n= (6-4)= )-(k+ 1) -(k+ 1) k+1 , we will prove that (6-4)= j=1 By inductive hypothesis
Use induction to prove: for any integer n ≥ 1,(6-4)= 3n² - n. Base case n = Inductive step Assume that for any k > k+1 Σ(6-4)= (6j - 4)+ j=1 j=1 = Σ(6; – 4) = j=1 = =( = 3 k² + k² + k+ k+ j=1 j=1 ,3n²-n= (6-4)= )-(k+ 1) -(k+ 1) k+1 , we will prove that (6-4)= j=1 By inductive hypothesis
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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