Ta đặt
\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+...+\dfrac{1}{49.51}=A\)
\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+..+\dfrac{2}{49.51}\)
\(2A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(\Rightarrow2A=\dfrac{1}{1}-\dfrac{1}{51}=\dfrac{50}{51}\)
\(A=\dfrac{50}{51}:2=\dfrac{25}{51}\)
Vậy : \(\dfrac{1}{3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}=\dfrac{25}{51}\)
Ta đặt \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(\Rightarrow A=\dfrac{1.2}{1.3.2}+\dfrac{1.2}{3.5.2}+\dfrac{1.2}{5.7.2}+...+\dfrac{1.2}{49.51.2}\)
\(\Rightarrow A=\dfrac{1}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(\Rightarrow A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(\Rightarrow A=\dfrac{1}{2}.\left(1-\dfrac{1}{51}\right)\)
\(\Rightarrow A=\dfrac{1}{2}.\dfrac{50}{51}\)
\(\Rightarrow A=\dfrac{25}{51}\)
