[Euler-Fermat]

1. $$\frac{1}{2}+\frac{1}{6}+....+\frac{1}{9900}$$
$$ = \frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}$$
$$ = [1-\frac{1}{2} +\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}]$$
$$ = [1-\frac{1}{100}] $$
$$ = \frac{99}{100} $$