Telescoping the Reciprocals!

Algebra Level 4

31!+2!+3!+42!+3!+4!+53!+4!+5!++10098!+99!+100!\begin{aligned} \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} + \cdots + \frac{100}{98!+99!+100!} \end{aligned}

Find the value of the expression above.

The answer is a form of 1a!1b!\dfrac{1}{a!} - \dfrac{1}{b!} , where aa and bb are integers. Submit your answer as a×ba \times b.

Notation: !! is the factorial notation. For example, 8!=1×2×3××88! = 1\times2\times3\times\cdots\times8 .

This problem is one of my set: Let's Practice.

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