Question

S=1/15+1/35+...+1/9999

Answer 1

\(S=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{9999}\)

\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{98}{303}\)

\(=\dfrac{49}{303}\)

Vậy \(S=\dfrac{49}{303}\)

#\(Toru\)