Answer:
<em>k^2 + k never stands for an odd integer</em>
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,
<em>Therefore, if k is an odd integer, then k^2 + k is an even integer
;</em>
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
<em>Therefore, if k is an even integer, then k^2 + k is an even integer;</em>
<em>This would make k^2 + k never stand for an odd integer</em>
