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Calculation of $\displaystyle\mathop{\sum\sum}_{1\leq i< j\leq n}ij$

When I expand that sum, I get

$\displaystyle[1(2+3+4+\cdots+n)+2(3+4+\cdots+n) $

$+3(4+5+6+\cdots+n)+\cdots (n-1)n]$

In each bracket sum in arthematic progression and we have to find individual sum

Is there is any better way to find its sum, have a look please.

Priti Bisht
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    Note that $$\underbrace{\sum_{1\leq i,j \leq n} ij}{=(\sum{i=1}^{n}i)^2} = \underbrace{\sum_{1\leq i=j\leq n} ij}{=\sum{i=1}^{n}i^2} + 2\left( \sum_{1\leq i<j\leq n}ij \right). $$ So, a formula for the sum can be easily derived from that for $\sum_{i=1}^{n}i$ and $\sum_{i=1}^{n}i^2$. – Sangchul Lee Jan 24 '23 at 09:14
  • Thanks Sangchul Lee got it. Can we say same about $\displaystyle \mathop{\sum\sum\sum}{1\leq <i<j\leq n}ijk=\frac{1}{3}[\sum{1\leq i,j,k\leq n}ijk-(\mathop{\sum}_{1\leq i=j=k\leq n}ijk)]$ – Priti Bisht Jan 24 '23 at 09:29
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    Unfortunately, no. In general you will need a suitable combination of sums over the ranges $$ 1\leq i,j,k\leq n, \qquad 1\leq i=j<k\leq n, \qquad 1\leq i<j=k\leq n, \qquad 1\leq i=j=k\leq n. $$ So, the trick quickly goes out of hand for higher-order versions. – Sangchul Lee Jan 24 '23 at 15:02
  • Thanks Sangchul Lee – Priti Bisht Jan 26 '23 at 06:09

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