# Telescoping the Reciprocals!

Algebra Level 4

$\begin{eqnarray} \frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \frac{5}{3!+4!+5!} + \cdots + \frac{100}{98!+99!+100!} \end{eqnarray}$

Find the value of the expression above.

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

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

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