Confusion on Summation Notation in Lagrange's Identity

Question

I'm working on a proof (the proof of Lagrange's Identity), but it includes a sum notation I'm not familiar with:

$$\sum_{1\le k\lt j\le n} (a_kb_j-a_jb_k)^2$$

I would appreciate any explanations of what this is saying, specifically in regards to the inequalities below the sigma. Thanks!

Answer 1

The notation is indeed confusing. It should be $$\sum_{\substack{k,j \\1\leq k<j\leq n}}$$ i.e., summing over $k$ and $j$ with the condition given below.

Answer 2

This means you are summing the value of $(a_kb_j-a_jb_k)^2$ over all ordered pairs $(k,j)$ such that $1\leq k<j\leq n$. For instance, if $n=3$, these ordered pairs are $(1,2)$, $(1,3)$, and $(2,3)$, so the sum is $$(a_1b_2-a_2b_1)^2+(a_1b_3-a_3b_1)^2+(a_2b_3-a_3b_2)^2.$$

Answer 3

For example, $1 \le k < j \le 5$ would indicate the $(j,k)$ pairs

\begin{array}{cccc} (2,1) \\ (3,1) & (3,2) \\ (4,1) & (4,2) & (4,3) \\ (5,1) & (5,2) & (5,3) & (5,4) \end{array}

Answer 4

${1\leq j<k\leq n}$ is the domain of the operator.   We sum the terms for all integer values of the bound variables, $(j,k)$, where this domain holds true.

This is sometimes more convenient than the double sum notation with which you might be more familiar. $$\sum_{1\leq j<k\leq n} (a_k b_j −a_j b_k )^2 \\ ~=~ \sum_{j=1}^{n-1}\sum_{k=j+1}^{n}(a_k b_j −a_j b_k )^2 \\ ~=~ \sum_{k=2}^{n}\sum_{j=1}^{k-1}(a_k b_j −a_j b_k )^2 $$