I'm totally lost. All I know is it has to do with binary trees and may need to be solved using induction.
Show that every 2-tree with $n$ internal nodes has $n + 1$ external nodes.
Show that the external path length ($epl$) in a 2-tree with $m$ external nodes satisfies $epl \leq \frac{1}{2}(m^2 + m - 2)$. Conclude that $epl \leq \frac{1}{2}n(n + 3)$ for a 2-tree with $n$ internal nodes.
Edit
In response to answer 2:
I know that if there are n internal nodes, there are n + 1 external (or m external and m - 1 internal) and that a tree has the maximum external path length if each node has at most one internal child. Such a length is defined by $epl_{max} = \frac{1}{2}n(n + 3)$. How do you go from $epl_{max}$ to $\frac{1}{2}m(m - 1) + (m - 1)$? And if $epl = n + 2m$, wouldn't it be,
$n + 2m <= \frac{1}{2}(m^2 + m - 2)$?
Ans 1:
For $n=1$, we have $0$ internal nodes and $1$ external node.
Induction Hypothesis: The 2-tree with $n-1$ internal nodes has $n$ external nodes.
Now, if you add $2$ nodes to one of the external node of a 2-tree with $n-1$ internal nodes, we get a 2-tree with $n$ internal nodes and the number of external nodes decreases by $1$ (since by adding $2$ nodes to an external node makes that (external) node an internal node in the new tree) and increases by 2 (since we are adding $2$ new nodes). This implies that the number of external nodes becomes $n-1+2 = n+1$.
Ans 2:
For a fixed number of external nodes, maximum $epl$ can be obtained when we make a tree which looks like this:
Red - Internal Nodes
Green - External Nodes
$epl \leq epl_{max} = \frac{1}{2}m(m-1) + (m-1) = (m-1)(\frac{m}{2} +1) = \frac{1}{2}(m-1)(m+2) = \frac{1}{2}(m^2 + m - 2)$
$m$ external nodes means $m-1$ internal nodes.
$n = m-1$. Put $m=n+1$ to get the desired result.