Tổng quát: $$\frac{1}{n(n+2)}=\frac{1}{2} \left( \frac{1}{n}-\frac{1}{n+2} \right )$$
Do đó: \(A=\frac{1}{3.5}+\frac{1}{5.7}+..+\frac{1}{97.99}\)
\(=\frac{1}{2} \left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+..+\frac{1}{95}-\frac{1}{97}+\frac{1}{97}-\frac{1}{99}\right) \)
\(=\frac{1}{2} \left(\frac{1}{3}-\frac{1}{99}\right) \)
\(=\frac{1}{2} \left( \frac{32}{99} \right) \)
\(= \frac{16}{99} \)
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