Lagrange Identity Sum Notation

Hi, how do I interpret the last sum: [Broken]

Sum (...)
1<=k < j <= n

Is it the double sum:

Sum( Sum( (a_k*b_j - a_j*b_k)^2 from k = 1 to n) from j = 2 to n ) ?
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In your notation it would be:

Sum( Sum( (a_k*b_j - a_j*b_k)^2 from k = 1 to j-1) from j = 2 to n )

as it's a sum of pairs (k,j) from {1,...,n}x{1,...,n} where k is strictly less than j.

If you arrange the pairs of {1,...,n}x{1,...,n} in an nxn grid with k indexing the rows and j the columns like so:

(1,1), (1,2), (1,3),...
(2,1), (2,2), (2,3),...

your sum is over the terms above the main diagonal, e.g. for n=3 it's the bold terms:

(1,1), (1,2), (1,3)
(2,1), (2,2), (2,3)
(3,1), (3,2), (3,3)


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I like to think of it as the sum over all vectors (k,i) where k and i can take any value btw 1 and n, BUT we consider only case vectors for which k<i.
Thanks, I see what the notation means, and your way of visualizing the index pairs of the terms being summed was very helpful too, as I never thought to think about it like that.

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