Question

G=1/3+1/15+.........+1/9999

Answer 1

\(G=\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\)

\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)

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

\(\Leftrightarrow G=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(\Leftrightarrow G=\frac{1}{2}.\frac{100}{101}\)

\(\Leftrightarrow G=\frac{50}{101}\)

Vậy : \(G=\frac{50}{101}\)