26 tháng 9 2023 lúc 19:40
\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)
\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{98}{303}\)
\(=\dfrac{49}{303}\)
Vậy \(S=\dfrac{49}{303}\)
#\(Toru\)
Đúng 4
Bình luận (0)
