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

\(=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{99\cdot101}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}=1-\frac{1}{101}=\frac{100}{101}\)

\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+.....+\frac{1}{9999}=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{99.}\)\(\frac{1}{99.101}\)

                                                            \(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{99}-\frac{1}{101}\)

                                                              \(=1-\frac{1}{101}=\frac{100}{101}\)

\(S=1:3+1:15+1:35+...+1:9999\)

\(S=\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{9999}\)

\(S=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)

\(2S=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)

\(2S=1-\frac{1}{101}\)

\(2S=\frac{100}{101}\)

\(S=\frac{100}{101}:2\)

\(S=\frac{50}{101}\)

1/3+1/15+1/35+1/63+1/99+……+1/9999
=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)
=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)
=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)
=1/2(1-1/101)
=1/2×(100/101)
=50/101

1/3+1/15+1/35+1/63+1/99+……+1/9999

=1/(1×3)+1/(3×5)+1/(5×7)+1/(7×9)+1/(9×11)+……+1/(99×101)

=1/2(1-1/3)+1/2(1/3-1/5)+1/2(1/5-1/7)+1/2(1/7-1/9)+1/2(1/9-1/11)+……+1/2(1/99-1/101)

=1/2(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9-1/11+……+1/99-1/101)

=1/2(1-1/101)

=1/2×(100/101)

=50/101 

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

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

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

= \(1.50-\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\) 

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

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

= \(50-\frac{1}{2}.\frac{100}{101}\)

= \(50-\frac{50}{101}\)

= \(\frac{5000}{101}\)