$\frac{1}{3} + \frac{1}{15} + \frac{1}{35} + \frac{1}{63} + ...... + \frac{1}{9999}$

= $\frac{1}{1 \times 3} + \frac{1}{3 \times 5} + \frac{1}{5 \times 7} + \frac{1}{7 \times 9} + ...... + \frac{1}{99 \times 101}$

= $\frac{1}{2} \times (1 - \frac{1}{3}) + \frac{1}{2} \times (\frac{1}{3} - \frac{1}{5}) + \frac{1}{2} \times (\frac{1}{5} - \frac{1}{7}) + \frac{1}{2} \times (\frac{1}{7} - \frac{1}{9}) ...... + \frac{1}{2} \times (\frac{1}{99} - \frac{1}{101})$

= $\frac{1}{2} \times [(1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + (\frac{1}{5} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{9}) ...... + (\frac{1}{99} - \frac{1}{101})]$

= $\frac{1}{2} \times (1 - \frac{1}{3} + \frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} ...... + \frac{1}{99} - \frac{1}{101})$

= $\frac{1}{2} \times (1 - \frac{1}{101})$

= $\frac{1}{2} \times \frac{100}{101}$

= $\frac{50}{101}$ .

2022-2023学年苏科版数学七上尖子生考点培优专题训练1 探索数与式的关系