Prove: For any integer n, the number (n2-n) is even. Use Proof by Cases to consider two cases where n is even and n is odd. Prove: For any integer n, the number (n²-n) is even. We will consider 2 cases where n is even andn is odd.Fill in the blanksi.а.Case 1: If n is even, then n=2k (k E Z).1.n2-n = (2k)2-(2k) = 4k2-2k = 2(2k2-k)2.Since 2k2-k is anwe can say that n²-n isi.Case 2: if n is odd, then n=2k-1 (k E Z).1. n2-n = (2k-1)2-(2k-1) = (4k2-4k+1)-(2k-1) = 2(2k2-3k+1)is an integer, we can say that n2-n is%3D%3D2. Since

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Prove: For any integer n, the number (n²-n) is even. We will consider 2 cases where n is even and n is odd. Fill in the blanks i. Case 1: If n is even, then n=2k (k E Z). а. n2-n = (2k)2-(2k) = 4k2-2k = 2(2k2-k) Since 2k2-k is an, 1. 2. , we can say that n²-n is ii. Case 2: if n is odd, then n=2k-1 (k E Z). 1. n2-n = (2k-1)2-(2k-1) = (4k2-4k+1)-(2k-1) = 2(2k2-3k+1) %3D 2. Since is an integer, we can say that n2-n is_

Prove: For any integer n, the number (n²-n) is even. We will consider 2 cases where n is even and n is odd. Fill in the blanks i. Case 1: If n is even, then n=2k (k E Z). а. n2-n = (2k)2-(2k) = 4k2-2k = 2(2k2-k) Since 2k2-k is an, 1. 2. , we can say that n²-n is ii. Case 2: if n is odd, then n=2k-1 (k E Z). 1. n2-n = (2k-1)2-(2k-1) = (4k2-4k+1)-(2k-1) = 2(2k2-3k+1) %3D 2. Since is an integer, we can say that n2-n is_

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Question

Prove: For any integer n, the number (n2-n) is even. Use Proof by Cases to consider two cases where n is even and n is odd.

Prove: For any integer n, the number (n²-n) is even. We will consider 2 cases where n is even and
n is odd.
Fill in the blanks
i.
а.
Case 1: If n is even, then n=2k (k E Z).
1.
n2-n = (2k)2-(2k) = 4k2-2k = 2(2k2-k)
2.
Since 2k2-k is an
we can say that n²-n is
i.
Case 2: if n is odd, then n=2k-1 (k E Z).
1. n2-n = (2k-1)2-(2k-1) = (4k2-4k+1)-(2k-1) = 2(2k2-3k+1)
is an integer, we can say that n2-n is
%3D
%3D
2. Since

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