\(A=\frac{4}{3}.\frac{4}{7}+\frac{4}{7}.\frac{4}{11}+\frac{4}{11}.\frac{4}{15}+...+\frac{4}{95}.\frac{4}{99}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{99}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{99}\)
\(\Leftrightarrow\frac{32}{99}\)
Đề bài sai đúng k??Kiểm tra lại đi nhé!!
Trịnh Thành Công làm sai rồi kìa
\(A=\frac{4}{3}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{4}{11}+...+\frac{4}{95}\cdot\frac{4}{99}\)
\(=\frac{4.4}{3.7}+\frac{4.4}{7.11}+...+\frac{4.4}{95.99}\)
\(=4\left(\frac{4}{3.7}+\frac{4}{7.11}+...+\frac{4}{95.99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{99}\right)\)
\(=4\left(\frac{1}{3}-\frac{1}{99}\right)=4\cdot\frac{32}{99}=\frac{128}{99}\)
