product of two consecutive integers is even

Can someone prove for me that n(n + 1) where n is an arbitrarily chosen integer can be represented as 2k where k is an integer? It was on the board today, and apparently I need to know it, but I was sleeping.

>>2 and >>4 have already shown what was required but for something different let
S = n + n-1 + n-2 + . . . + 1
S = 1 + 2   + 3   + . . . + n
S is an element of the integers since integers are closed under addtion.  Adding related components gives
2S = (n+1) + (n+1) + (n+1) + . . . + (n+1)
   = n(n+1)
=> n(n+1) is an even integer since 2*integer is always even.