Question

tính tổng \(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\) 

Answer 1

\(M=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{9603}+\dfrac{2}{9999}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{6}{5}+\dfrac{3}{2}\cdot\dfrac{96}{505}=\dfrac{150}{101}\)