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{align*} \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{align*}and so on. The sum of the first $ n $ terms appears to be $ \frac{n}{n + 1} $. Suppose that \[\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{(n - 1)n} + \frac{1}{n(n + 1)} = \frac{n}{n + 1} = 1 - \frac{1}{n + 1}.\]Then \[\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{(n - 1)n} = \frac{n - 1}{n} = 1 - \frac{1}{n}.\]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{align*} \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) + \dots + \left( \frac{1}{99} - \frac{1}{100} \right) \\ &= 1 - \frac{1}{100} = \boxed{\frac{99}{100}}.\end{align*}