Triangle measures

Let $T_n$ be the number of all triples $a,b,c$ of integers with $1\leq a \leq b \leq c \leq 2n$ that can make the side lengths of a triangle (which simply means that $a+b>c$ ). Examining the cases $c=2n+1$ and $c=2n+2$ and doing a little counting, we find the recursion $T_{n+1}=n(2n-1)+T_{n}$ , so that $T_{50}=\sum_{n=1}^{49}n(2n-1)=49\times 50(33-0.5)$ . Expressed as a proportion of all ordered triples, ${100}\choose {3}$ , the answer is $\frac{49\times 50\times 195}{100\times 99 \times 98}=\boxed{\frac{65}{132}}$ .