To prove 3+7+11+...+(4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, under our previous assumption, we prove that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. OP(n+1): 3+7+11+. . +4(n+1)-1= 2n² + 5n +31 P(k+1): 3+7+11+...+ 4(k+ 1) − 1 = (k+1)(2(k+1)+1) n+1 P(n+1): (4(+1) − 1) = (n + 1)(2(n + 1) + 1) j=1 n+1 P(n+1): (4j-1) = (n + 1)(2(n+1) + 1) Σw-

To prove 3+7+11+...+(4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, under our previous assumption, we prove that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. OP(n+1): 3+7+11+. . +4(n+1)-1= 2n² + 5n +31 P(k+1): 3+7+11+...+ 4(k+ 1) − 1 = (k+1)(2(k+1)+1) n+1 P(n+1): (4(+1) − 1) = (n + 1)(2(n + 1) + 1) j=1 n+1 P(n+1): (4j-1) = (n + 1)(2(n+1) + 1) Σw-

Elements Of Modern Algebra

8th Edition

ISBN:9781285463230

Author:Gilbert, Linda, Jimmie

Publisher:Gilbert, Linda, Jimmie

Chapter2: The Integers

Section2.2: Mathematical Induction

Problem 49E: Show that if the statement is assumed to be true for , then it can be proved to be true for . Is...

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