Umm... I can't remember the exact formula for triangular series, but I know how to solve them without one. Help!

The n'th triangular number is given by n(n+1)/2

The sum of the first n triangular numbers is n(n+1)(n+2)/6

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I'm not good with this.....but I bet Alan could help.....either you can wait for him or shoot him a message....he's great at this for what I've seen!

See?  I told you!

Yes, I suspected that the formula might be based off the area of any triangle whose perpendicular height is known (1/2*bh)

Anyway, Thanks