Câu trả lời:

\(A=\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{99.101}\)

\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-...-\frac{1}{101}\right)=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{101}\right)\)

\(A=\frac{49}{303}\)