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!
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!
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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
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