Question

\(\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+.....+\frac{4}{95}\cdot\frac{4}{99}\)

Answer 1

Đặt \(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+...+\frac{4}{95}\cdot\frac{4}{99}\)

\(A=\frac{16}{21}+\frac{16}{77}+...+\frac{16}{9405}\)

\(A=\frac{16}{3\cdot7}+\frac{16}{7\cdot11}+....+\frac{16}{95\cdot99}\)

\(A=\frac{16}{4}\cdot\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(A=4\cdot\left(\frac{1}{3}\cdot\frac{1}{99}\right)=4\cdot\frac{32}{99}=\frac{128}{99}\)