The length of the path from $v_5$ to $v_6$ in the MST of previous question with $n = 10$ is

From the previous question, the edges present in the MST are:

1. Edge $(v_1, v_2)$ with weight $3$.

2. Edges $(v_i, v_{i+2})$ with weight $i + (i+2) = 2i + 2$ for all $1 \le i \le n-2$.

In the MST, the vertices are connected in two chains through alternating nodes linked together at $v_1$ and $v_2$:

• The odd chain: $v_5 - v_3 - v_1$

• The edge between chains: $v_1 - v_2$

• The even chain: $v_2 - v_4 - v_6$

Thus, the unique simple path from $v_5$ to $v_6$ in the MST is:

The edges on this path and their weights are:

• $(v_5, v_3)$: weight $= 5 + 3 = 8$

• $(v_3, v_1)$: weight $= 3 + 1 = 4$

• $(v_1, v_2)$: weight $= 1 + 2 = 3$

• $(v_2, v_4)$: weight $= 2 + 4 = 6$

• $(v_4, v_6)$: weight $= 4 + 6 = 10$

Total length of the path:

Therefore, option C is the correct choice.