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_

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