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, \(8! = 1\times2\times3\times\cdots\times8 \).


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