Question

Tính E=\(\frac{2}{3}+\frac{14}{15}+\frac{34}{35}+...+\frac{9998}{9999}\)

Answer 1

\(E=\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{15}\right)+...+\left(1-\frac{1}{9999}\right)\)

    \(=\left(1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\right)\)

     \(=50-\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\right)\)

     \(=50-\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)

     \(=50-\frac{1}{2}.\left(1-\frac{1}{101}\right)=50-\frac{1}{2}.\frac{100}{101}=50-\frac{50}{101}=\frac{5000}{101}\)