Lời giải:
Ta có: \(G=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}\)
\(\Rightarrow2.G=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+.....+\frac{2}{9999}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
\(\Rightarrow G=\frac{50}{101}\) . Vậy: \(\\G=\frac{50}{101}\)
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\(G=\frac{1}{3}+\frac{1}{15}+...+\frac{1}{9999}\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+....+\frac{1}{99.101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(\Leftrightarrow G=\frac{1}{2}.\frac{100}{101}\)
\(\Leftrightarrow G=\frac{50}{101}\)
Vậy : \(G=\frac{50}{101}\)