jmc

First of all, the denominators of these fractions are sometimes called ``oblong numbers'' since they make rectangles that are one unit longer than they are wide: $1\times 2=2,2\times 3=6,3\times 4=12,4\times 5=20$, etc. The last denominator in the expression is $99\times 100=9900$. Let's find the sum of a few terms at a time and see if we notice a pattern. $$\begin{array}{rl}\frac{1}{2}+\frac{1}{6}& =\frac{2}{3},\\ \frac{1}{2}+\frac{1}{6}+\frac{1}{12}& =\frac{3}{4},\\ \frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}& =\frac{4}{5},\end{array}$$and so on. The sum of the first $n$ terms appears to be $\frac{n}{n+1}.$

Suppose that $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots +\frac{1}{(n-1)n}+\frac{1}{n(n+1)}=\frac{n}{n+1}=1-\frac{1}{n+1}.$$Then $$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots +\frac{1}{(n-1)n}=\frac{n-1}{n}=1-\frac{1}{n}.$$Subtracting these equations, we obtain $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}.$$Note that we can algebraically verify this identity: $$\frac{1}{n}-\frac{1}{n+1}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n(n+1)}.$$Therefore, the sum of the 99 fractions in the expression is $$\begin{array}{rl}\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\cdots +\frac{1}{n(n+1)}+\cdots +\frac{1}{9900}& =\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\cdots +\left(\frac{1}{99}-\frac{1}{100}\right)\\ & =1-\frac{1}{100}=\boxed{\frac{99}{100}}.\end{array}$$