Consider a weighted complete graph \(𝐺\) on the vertex set…

Answer: 4n - 4

Reasoning:

• Edge weights are given by 4|i - j|, so any edge joining consecutive vertices vi and vi+1 has weight 4.

• A spanning tree on n vertices uses exactly n-1 edges, so using the n-1 consecutive edges (v1–v2, v2–v3, …, v_{n-1}–v_n) gives total weight 4 + 4 + … + 4 = 4(n-1).

• To see this is minimal, consider any cut that separates the first k vertices {v1,…,vk} from the rest. The smallest edge crossing that cut is between vk and vk+1 and has weight 4, so any spanning tree must include at least one crossing edge of weight at least 4 for each such cut. There are n-1 such cuts, giving a lower bound of 4(n-1).

• Since the chain of consecutive vertices attains this lower bound, the minimum possible spanning tree weight is 4(n-1) = 4n - 4.