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 \(\frac{1}{a!} - \frac{1}{b!} \). 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.
by
Fidel Simanjuntak