có ai muốn kb ko fa quá trời
\(A=\frac{1}{1.2}+\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{47.49}\)
\(\Rightarrow2A=1+\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{47.49}\)
ta có: \(\frac{2}{1.3}=1-\frac{1}{3}\)
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
...
\(\frac{2}{47.49}=\frac{1}{47}-\frac{1}{49}\)
vậy \(2A=1+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{45}-\frac{1}{47}+\frac{1}{47}-\frac{1}{49}\)
\(\Rightarrow2A=2-\frac{1}{49}\)
\(\Rightarrow2A=\frac{97}{49}\)
\(\Rightarrow A=\frac{97}{98}\)