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Write as a single sum:

Given $\{a_n\}_n$, $a_i \in \mathbb{Q}$, $0 \lt a_i \le a_{i+1}$

$\sum_{i=1}^{n} \sum_{j=i+1}^{n} \lfloor a_j - a_i \rfloor$

I am not sure if this is possible.

I know that if there is no floor, then the sum can be written as:

$\sum_{i=1}^{n} (2i-n-1) a_i$

1 Answers 1