Use induction to show that
Assume is true.
Now what?
K or k+1 is even.
Any amount of an even value is even.
Via induction...
P(k)
k(k+1) is even ?
P(k+1)
(k+1)(k+2) is even ?
Try to prove that P(k) being true also causes P(k+1) to be true
Proof
If P(k) is true, then P(k+1) is true as 2(k+1) is a multiple of 2 and so is even.