Telescoping the Reciprocals!

$\begin{aligned} S & = \frac 3{1!+2!+3!} + \frac 4{2!+3!+4!} + \frac 5{3!+4!+5!} + \cdots + \frac {100}{98!+99!+100!} \\ & = \sum_{n=1}^{98} \frac {n+2}{n!+(n+1)!+(n+2)!} \\ & = \sum_{n=1}^{98} \frac {{\color{#3D99F6}(n+1)}(n+2)}{n!{\color{#3D99F6}(n+1)}+{\color{#3D99F6}(n+1)}(n+1)!+{\color{#3D99F6}(n+1)}(n+2)!} \\ & = \sum_{n=1}^{98} \frac {(n+1)(n+2)}{(n+1)!(1+n+1)+(n+1)(n+2)!} \\ & = \sum_{n=1}^{98} \frac {(n+1)(n+2)}{(n+2)(n+2)!} \\ & = \sum_{n=1}^{98} \frac {n+1}{(n+2)!} \\ & = \sum_{n=1}^{98} \frac {n+2-1}{(n+2)!} \\ & = \sum_{n=1}^{98} \left( \frac 1{(n+1)!} - \frac 1{(n+2)!} \right) \\ & = \frac 1{2!} - \frac 1{100!} \end{aligned}$

$\implies a \times b = 2 \times 100 = \boxed{200}$

Now, note that $\frac{3}{1!+2!+3!} = \frac{3}{1!(1+2+2 \times 3)}$

$\frac{3}{1! \times 9} = \frac{1}{3} = \frac{2}{3 \times 2 \times 1}$

$= \frac{2}{3!} = \frac{1}{2!} - \frac{1}{3!}$ .

We have $\frac{3}{1!+2!+3!} = \frac{1}{2!} - \frac{1}{3!}$

Simplify each terms, we have

$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} + ... + \frac{100}{98!+99!+100!}$

$= \frac{1}{2!} - \frac{1}{3!} + \frac{1}{3!} - \frac{1}{4!} + \frac{1}{4!} - \frac{1}{5!} + ........ - \frac{1}{100!}$

After all, we have

$\frac{1}{2!} - \frac{1}{100!} = \frac{1}{a!} - \frac{1}{b!}$

$a=2$ and $b=100$ .

Hence, $a \times b = \boxed{200}$ .