To prove 3+ 7+ 11 + ... + (4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, we first assume that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. 77 P(k) : Σ(4k - 1) = n(2n + 1) k=1 72 P(n): (4n-1) = n(2n +1) j=1 72 P(n): (4j-1) = n(2n +1) j=1 n+1 P(n + 1): (4j-1) = (n + 1)(2(n + 1) + 1) j=1

To prove 3+ 7+ 11 + ... + (4n − 1) = n(2n + 1) holds for an arbitrary n € Z+, in the inductive step, we first assume that Check ALL that apply, i.e. you are allowed to choose ONE or MORE answers. 77 P(k) : Σ(4k - 1) = n(2n + 1) k=1 72 P(n): (4n-1) = n(2n +1) j=1 72 P(n): (4j-1) = n(2n +1) j=1 n+1 P(n + 1): (4j-1) = (n + 1)(2(n + 1) + 1) j=1

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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Show that if the statement is assumed to be true for , then it can be proved to be true for . Is the statement true for all positive integers ? Why?
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