Another Pattern

$\begin{aligned} S & = \frac 1{1\times 2\times 3} + \frac 1{2\times 3\times 4} + \frac 1{3\times 4\times 5} + ... + \frac 1{98\times 99\times 100} \\ & = \sum_{n=1}^{98} \frac 1{n(n+1)(n+2)} \quad \quad \small \color{#3D99F6} \text{Using partial fractions} \\ & = \frac 12 \sum_{n=1}^{98} \left( \frac 1n - \frac 2{n+1} + \frac 1{n+2} \right) \\ & = \frac 12 \sum_{n=1}^{98} \left( \frac 1n - \frac 1{n+1} - \frac 1{n+1} + \frac 1{n+2} \right) \quad \quad \small \color{#3D99F6} \text{By telescoping} \\ & = \frac 12 \left( \frac 11 - \frac 1{99} - \frac 12 + \frac 1{100} \right) \\ & = \frac {4949}{19800} \\ & = \frac {4950-1}{19800} \quad \quad \small \color{#3D99F6} \text{Multiply up and down by 4} \\ & = \frac {19800-4}{4(19800)} \quad \quad \small \color{#3D99F6} \text{Note that } \frac 1p - \frac 1q = \frac {q-p}{pq} \\ & = \frac 14 - \frac 1{19800} \end{aligned}$

$\implies p+q = 4+19800 = \boxed{19804}$

First, let's determine the pattern.

$\frac{1}{1 \times 2 \times 3} = \frac{1}{6}$

$\frac{1}{1 \times 2 \times 3} + \frac{1}{2 \times 3 \times 4} = \frac{5}{24}$

Then, we have the pattern

$\frac{1}{1\times 2 \times 3} + \frac{1}{2 \times 3 \times 4} + ... + \frac{1}{(n-2)(n-1)(n)}$

$= \left( \frac{1}{2} - \frac{1}{(n)(n-1)} \right)\left( \frac{1}{2} \right)$ .

Now, put $n=100$ , we have

$\left( \frac{1}{2} - \frac{1}{9900} \right)\left( \frac{1}{2} \right)$

$= \frac{1}{4} - \frac{1}{19800} = \frac{1}{p} - \frac{1}{q}$

$p = 4$ and $q = 19800$

Hence, the answer is $p + q = \boxed{19804}$ .