Sure! A triangle number can be thought as the total object count when building a pyramid where each layer has one more item from the last, first a visual:

$N=1$)
  o
$T(N) = 1$

$N=2$)
  o
 o o
$T(N) = 3$

$N=3$)
  o
 o o
o o o
$T(N) = 6$


$N=4$)
   o
  o o
 o o o
o o o o
$T(N) = 10$

Steps:
1) Note that in each sum, adding row $1$ to row $N$ equals $N+1$, but so does adding row $2$ to row $N-1$. 
2) In even cases ($N=2*M$) we have a total of $N/2$ copies of $N+1$, or $N*(N+1)/2$.
3) In odd cases ($N=2*M+1$) we have $(N-1)/2$ copies of $N+1$, plus the middle row of $(N+1)/2$ for a total of $ (N-1)*(N+1)/2 + (N+1)/2 = N*(N+1)/2 $.
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