Simplify Multiplication of Fractions

[Albert1]

$\dfrac{3}{3\times 4}+\dfrac{4}{3\times 4\times 5}+\dfrac{5}{3\times 4\times 5\times 6}+\cdots+\dfrac {99}{3\times 4\times 5\times 6\times \cdots\times 99\times 100}$

 

[alyafey22]

Re: count value

$\dfrac{3}{3\times 4}+\dfrac{4}{3\times 4\times 5}+\dfrac{5}{3\times 4\times 5\times 6}+ \cdots +\dfrac {99}{3\times 4\times 5\times 6\times \cdots \times 99\times 100}$

$=2 \left( \dfrac{3}{2 \times 3\times 4}+\dfrac{4}{2\times 3\times 4\times 5}+\dfrac{5}{2\times 3\times 4\times 5\times 6}+ \cdots +\dfrac {99}{2 \times 3\times 4\times 5\times 6\times \cdots \times 99\times 100} \right)$

\(\displaystyle =2 \sum^{100}_ {n=4} \frac{n-1}{n!}=2 \sum^{100}_ {n=4} \frac{1}{(n-1)!}-\frac{1}{n!} \)

The sum is telescoping

\(\displaystyle 2\left( \dfrac{1}{3!} -\dfrac{1}{4!} + \dfrac{1}{4!} - \dfrac{1}{5!} + \frac{1}{5!}-\frac{1}{6!}+ \cdots +\dfrac {1}{99!} - \dfrac {1}{100!} \right)= 2 \left(\dfrac{1}{3!}-\dfrac{1}{100!} \right) \)