Show that the diameter of a MST is sometimes larger by a factor $\Omega(n)$ than the diameter of the graph $G$
You say that you don't understand how to show that the diameter of a MST can be $\Omega(n)$ larger than the diameter of the underlying graph.

However, it seems that what you don't understand is what it means that the diameter of a MST can be $\Omega(n)$ larger than the diameter of the underlying graph.

It is impossible to prove a theorem if you don't understand its statement.

Here is what you are asked to do. Given a weighted graph $G$, we will use the following notation:


$n(G)$ is the number of vertices in $G$.
$D(G)$ is the diameter of $G$.
$D_{MST}(G)$ is the minimum diameter of a minimum spanning tree of $G$.



  Give a sequence of graphs $G_i$ such that $n(G_i) \to \infty$ and $D_{MST}(G) = \Omega(n(G)) \cdot D(G)$.


It seems that you are not familiar with the notation $\Omega(\cdot)$. It is the counterpart of big O, and described in the answers to this question.

Unpacking the big $\Omega$ notation, we can rephrase the question:


  Give a sequence of graphs $G_i$ such that $n(G_i) \to \infty$ and there exists a constant $C&gt;0$ such that $D_{MST}(G_i) \geq C n(G_i) D(G_i)$.


The question is somewhat ambiguous – it's not clear if you need to provide an example for every $n$ or just for infinitely many $n$. I assumed the latter, but in fact in your case there is a sequence $G_i$ with $n(G_i) = i$, so you can try proving this stronger statement instead.