Sum of all triangle numbers Does anyone know the sum of all triangle numbers? I.e 
1+3+6+10+15+21... 
I've tried everything, but it might help you if I tell you one useful discovery I've made: 
I know that the sum of alternating triangle numbers, 1-3+6-10... Is equal to 1/8 and that to change
1+3+6... Into 1-3+6... You would subtract 6+20+42+70... which is every other triangular number (not the hexagonals) multiplied by two. 
1/8 plus this value is 1+3+6+10+...
A final note: I tried to split the triangle numbers into hexagonals and that series and then I got the squares of the odd numbers. Using dirichlet lambda functions This gave me 0 but I don't think this could be right. A number of other sums gave me -1/24 and 3/8 but I have no idea

The $r$-th triangular number is 
$$T_r=\frac {r(r+1)}2=\binom {r+1}2$$
 i.e. $1, 3,6, 10, ...$ for $r=1, 2, 3, 4, ...$.
The sum of the first $n$ triangular numbers is
$$S_n=\sum_{r=1}^n T_r=\color{blue}{\sum_{r=1}^n \binom {r+1}2=\binom {n+2}3}=\frac {(n+2)(n+1)n}6$$
i.e. $1, 4, 10, 20, ...$ for $n=1, 2, 3, 4...$. This is also known as the $n$-th tetrahedral number. 
