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You know this how you can express a sum of consecutive terms like this: $1 + 2 + 3 + ... + n = \frac{n(n + 1)}{2}$? In other words, only using $n$?

I would like to know if there is a way to shorten, or to express such a sum of sums:

For example, for $n = 4$ $$\sum_{i = 1}^{4}\frac{i(i + 1)}{2} =1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) = 20$$

Can I express this using only $n$?

davaisk
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  • you might be interested in discrete calculus, which gives you a way to compute closed form formulas for any polynomial recursive sequence, and probably more... –  Nov 23 '20 at 12:57
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    Looks like $(n^3+3n^2+2n)/6$ – Hagen von Eitzen Nov 23 '20 at 13:02
  • $\frac{i(i+1)}{2}=\binom{i+1}{2}$ is the number of ways to choose 3 positive integers, the largest of which is $i+2$. Therefore $\sum_{i=1}^{n}{\frac{i(i+1)}{2}}$ is the number of ways to choose 3 positive integers smaller than or equal to $n+2$ i.e. $\binom{n+2}{3}$ – acat3 Nov 23 '20 at 15:01

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