Đặt :
\(A=\dfrac{1}{3.5}+\dfrac{1}{5.7}+.........+\dfrac{1}{99.101}\)
\(\Leftrightarrow2A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+............+\dfrac{2}{99.101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+............+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{1}{3}-\dfrac{1}{101}\)
\(\Leftrightarrow2A=\dfrac{98}{303}\)
\(\Leftrightarrow A=\dfrac{49}{303}\)
\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+...+\dfrac{1}{99.101}\)
= \(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
= \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
= \(\dfrac{1}{2}.\dfrac{98}{303}=\dfrac{49}{303}\)
Gọi B là tổng của gái trị trên Suy ra:
B= \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+.....+\dfrac{1}{99.11}\)
B=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+....+\dfrac{1}{99}-\dfrac{1}{100}\)
B=\(\dfrac{1}{3}-\dfrac{1}{101}\)
B= \(\dfrac{101}{303}-\dfrac{3}{303}\)
B=\(\dfrac{98}{303}\)
