If k stands for an integer, then is it possible for k2 + k to stand for an odd integer? Be prepared to justify your answer.

k^2 + k never stands for an odd integer

Step-by-step explanation:

Let us consider either case with which k stands for an odd or even integer;

Case 1: k is an odd integer

For integer a, k = 2a + 1

So, k + 1 = 2a + 2 = 2 (a + 1) = 2b for integer b

k^2 + k = k (k + 1) = k (2b) = 2kb = 2c for integer c,

Therefore, if k is an odd integer, then k^2 + k is an even integer;

Case 2: k is an even integer

For an integer a, k = 2a

So, k + 1 = 2a + 1

k^2 + k = k (k+1) = 2a (2a + 1), multiple of 2

Therefore, if k is an even integer, then k^2 + k is an even integer;

This would make k^2 + k never stand for an odd integer