Infinite summation of n(n+1)
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Infinite summation of n(n+1)

[From: ] [author: ] [Date: 11-11-28] [Hit: ]
Thanks!-Are you summing from n=1 to infinity? Because in that case, this diverges- the sum is infinity.Otherwise, its the sum of the first n squares plus the sum of the first n numbers,......
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!

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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
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