$\begin{aligned}&\frac{1}{3×5} + \frac{1}{5×7} + \frac{1}{7×9} + \cdots + \frac{1}{97×99}\\=&\frac{1}{2}×(\frac{1}{3} - \frac{1}{5} + \frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + \cdots + \frac{1}{97} - \frac{1}{99})\\=&\frac{1}{2}×\frac{32}{99}\\=&\frac{16}{99}\end{aligned}$