Lets say you have two weighted graphs $G_1=(V_1, E_1, d_1), G_2=(V_2, E_2, d_2)$ which do not have common vertices. You would like to union $G_1$ and $G_2$ by adding new edges (for instance, by adding the one new edge only). You can denote the union by $$G_1 + G_2 = (V_1 \cup V_2, E_1 \cup E_2 \cup E' \cup V_1 \times V_2 \cup V_2 \times V_1, d_1 \cup d_2 \cup d'), $$ where $E'$ is the set of new (added) edges, and $d'$ is the set of weights corresponded to added edges and defined by some rules (for instance, all elements are constant).
Question. How to expand the formula on the common case when you have $n$ graphs: $G_1=(V_1, E_1, d_1), G_2=(V_2, E_2, d_2), \ldots, G_n=(V_n, E_n, d_n)$?
I have tried to write:
$$\bigcup_{i=1}^n G_i=G_1+G_2+\ldots G_n=\\(V_1 \cup V_2 \cup \ldots \cup V_n, \\E_1 \cup E_2 \cup \ldots E_n \cup E' \cup \color{red}{V_1 \times V_2 \times \ldots \times V_n \cup V_n \times V_{n-1} \times \ldots \times V_1}, \\d_1 \cup d_2 \cup \ldots \cup d_n \cup d').$$
I'm not sure in the red part $V_1 \times V_2 \times \ldots \times V_n \cup V_n \times V_{n-1} \times \ldots \times V_1$.