Ta có: \(A=\frac{1}{3}.\frac{1}{7}+\frac{1}{7}.\frac{1}{11}+\frac{1}{11}.\frac{1}{15}+...+\frac{1}{95}.\frac{1}{99}\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+.....+\frac{1}{95}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{99}\right)\)
\(=\frac{1}{4}.\frac{32}{99}=\frac{8}{99}\)
A = 1/4 x (1/3-1/7+1/7-1/11+1/11-1/15+..........+1/95-1/99)
= 1/4 x(1/3-1/99)
= 1/4 x 32/99
= 8/99
\(A=\frac{1}{3}\cdot\frac{1}{7}+\frac{1}{7}\cdot\frac{1}{11}+\frac{1}{11}\cdot\frac{1}{15}+...+\frac{1}{95}\cdot\frac{1}{99}\)
\(\Leftrightarrow A=\frac{1}{3\cdot7}+\frac{1}{7\cdot11}+\frac{1}{11\cdot15}+...+\frac{1}{95\cdot99}\)
\(\Leftrightarrow A=\left(\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{95\cdot99}\right)\div4\)
\(\Leftrightarrow A=\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\right)\cdot\frac{1}{4}\)
\(\Leftrightarrow A=\left(\frac{1}{3}-\frac{1}{99}\right)\cdot\frac{1}{4}\)
\(\Leftrightarrow A=\left(\frac{33}{99}-\frac{1}{99}\right)\cdot\frac{1}{4}\)
\(\Leftrightarrow A=\frac{32}{99}\cdot\frac{1}{4}\)
\(\Leftrightarrow A=\frac{32\cdot1}{99\cdot4}\)
\(\Leftrightarrow A=\frac{4\cdot8\cdot1}{99\cdot4}\)
\(\Leftrightarrow A=\frac{8}{99}\)
A=(1/3-1/7+1/7-1/11+.....+1/95-1/99)x1/4
A=(1/3-1/99)x1/4
A=1/4x32/99
A=8/99