Question

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

Giúp tui với :((((

Answer 1

\(M=1+\dfrac{1}{5}+\dfrac{3}{35}+...+\dfrac{3}{9603}+\dfrac{3}{9999}\)
     \(=\left(1+\dfrac{1}{5}\right)+\left(\dfrac{3}{5\cdot7}+...+\dfrac{3}{97\cdot99}+\dfrac{3}{99\cdot101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\left(\dfrac{1}{5}-\dfrac{1}{101}\right)\)
     \(=\dfrac{6}{5}+\dfrac{2}{3}\cdot\dfrac{96}{505}\)
     \(=\dfrac{6}{5}+\dfrac{64}{505}\)
     \(=\dfrac{134}{101}\)

\(\#PeaGea\)