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In our maths class we are currently studying sums, as an exercice we were given the task of proving the following : $$\sum_{1 \le j \lt k \le n} (a_k - a_j)(b_k-b_j) =\dfrac{1}{2}\sum_{1 \le j,k \le n} (a_k - a_j)(b_k-b_j)$$ Where $a_k , b_k$ are random reals.
I did some research and found people explaining these double sums using matrices but I can't understand them.
To be specific i'm looking for a more algebraic proof of this equality and grasping the meaning behind the double sum notations and the difference between the two sums shown above. Any help would be greatly appreciated!

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$\sum_{1\le j,k \le n} (a_k-a_j)(b_k-b_j)$ = $\sum_{1\le j \lt k \le n} (a_k-a_j)(b_k-b_j)$ + $\sum_{1\le k\lt j \le n} (a_k-a_j)(b_k-b_j)$ + $\sum_{1\le k=j \le n} (a_k-a_j)(b_k-b_j)$


$\sum_{1\le j \lt k \le n} (a_k-a_j)(b_k-b_j)$ = $\sum_{1\le k\lt j \le n} (a_k-a_j)(b_k-b_j)$

As $(a_k-a_j)(b_k-b_j)= (a_j-a_k)(b_j-b_k)$


$\sum_{1\le k=j \le n} (a_k-a_j)(b_k-b_j)$ = 0

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  • $\begingroup$ Exactly what I needed. Thank you! $\endgroup$ – OUCHNA Sep 20 '20 at 18:23
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Writing down explicitly an example for a small $n$ can help facilitate both the explanation of the notation as well as the proof.

For $n=3$,

\begin{align} \sum_{1 \le j < k \le n} (a_k - a_j)(b_k - b_j) &= (a_2 - a_1)(b_2 - b_1) + (a_3 - a_1)(b_3 - b_1) + (a_3 - a_2)(b_3 - b_2) \\ \sum_{1 \le j,k \le n} (a_k - a_j)(b_k - b_j) &= (a_1 - a_1)(b_1 - b_1) + (a_1 - a_2)(b_1 - b_2) + (a_1 - a_3)(b_1 - b_3) \\ &\qquad + (a_2 - a_1)(b_2 - b_1) + (a_2 - a_2)(b_2 - b_2) + (a_2 - a_3)(b_2 - b_3) \\ &\qquad + (a_3 - a_1)(b_3 - b_1) + (a_3 - a_2)(b_3 - b_2) + (a_3 - a_3)(b_3 - b_3) \end{align}

The difference in the two summation notations is that the first one only involves terms where $k > j$, while the second line considers all pairs $(j,k)$. Verifying the equation for $n=3$ should be fairly straightforward from here, and you can generalize to arbitrary $n$.

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  • $\begingroup$ Thanks for your input! $\endgroup$ – OUCHNA Sep 20 '20 at 18:24

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