2^2 + 2 = 4 +2 = 6 again... yes it does ..
etc, etc ..
So: the induction proof :
lets assume it works for some n : Then (n^2 +n) /2 evenly !
now try it for n ---> n+1
( by subbing in (n+1 ) for the n )
then for n---> n+1 we get :
(n+1)^2 + (n+1 ) = n^2 +2n +1 + n +1
= n^2 +3n + 2
Now we break THIS apart, by splitting up the 3n into two parts :
n^2 +3n + 2 = [ n^2 +n] + [2n +2 ]
Taking it term by term:
[ n^2 + n ] does divide by 2 ( original assumption of the induction proof)
and for the other term, [ 2n+2 ] this factors into 2[n+1 ] which also divides by 2 ( at least the '2' part does. )
so both terms divide by 2
thus if it works for n, then it works for n+1...
as in :
it works for 2 ( Vide supra ) so it works for 3
It works for 3 so it works for 4
It works for 4 so it works for 5 ..... etc , etc, etc up to infinity ..
QED