Infinite summation of n(n+1)

Title says it all. How would I find the summation? Really confused.

So it's like 2 + 6 + 12 ... + n(n+1) = ?

I took the integral of the Reimann sum, but it's not too accurate.

Thanks!

Are you summing from n=1 to infinity? Because in that case, this diverges- the sum is infinity.

Otherwise, it's the sum of the first n squares plus the sum of the first n numbers, both of which have formulas:

sum (n(n+1)) = sum(n^2) + sum(n) = n(n+1)(2n+1)/6 + n(n+1)/2