General formula for $\prod_{i<j} (a_i + b_j)$

Question

I want to expand the product of a sum into a sum of products $$ \prod_{i<j}^n (a_i + b_j) = \sum_{\text{sets } A,B} ~ \prod_{i\in A} a_i \prod_{j\in B} b_j. $$

---

With the result from this post General formula for $\prod (x+a_i)$, it follows that $$ \prod_{i<j} (a_i + b_j) = \prod_{j=2}^n \prod_{i=1}^{j-1} (a_i + b_j) = \prod_{j=2}^n \Big( \sum_{A \subset \{1, .., j-1\}} b_j^{j-1-|A|} \prod_{i\in A} a_i \Big). $$ However, I don't see how to continue from here.

There is also this post: expand the product $\prod_{i=1}^n(a_i+b_i)$, but again it does not help in this generalized case.

Answer 1

A representation which might be useful. We write the indices as set of pairs: \begin{align*} A:=\{(i,j)\in\mathbb{N}\times\mathbb{N}\big| 1\leq i<j\leq n\} \end{align*} We also assume wlog $b_j\neq 0, (2\leq j\leq n)$, since otherwise we can factor out all terms $a_k$ with $b_k=0$ and after reorder we have again something like the initial situation.

We obtain \begin{align*} \color{blue}{\prod_{1\leq i<j\leq n}\left(a_i+b_j\right)} &=\prod_{j=2}^nb_j^{j-1}\prod_{1\leq i<j\leq n}\left(1+\frac{a_i}{b_j}\right)\\ &=\prod_{j=2}^nb_j^{j-1}\prod_{(i,j)\in A}\left(1+\frac{a_i}{b_j}\right)\\ &\,\,\color{blue}{=\sum_{S\subseteq A}\prod_{s\in S}\frac{a_{\pi_{1}(s)}}{b_{\pi_{2}(s)}}\prod_{j=2}^nb_j^{j-1}} \end{align*}

In the last line we sum over all subsets of $S$ and use the projection operator \begin{align*} \pi_{1}(s)&=\pi_{1}\left(i,j\right)=i\\ \pi_{2}(s)&=\pi_{2}\left(i,j\right)=j \end{align*}