B=3/2x4/3x...........x2018/2017

=3x4x5x...........x2018/2x3x2x2x............x2017

=2x2018

=4036

A,C tương tự

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

\(=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{99\cdot101}\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\left(\dfrac{1}{3}-\dfrac{1}{101}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{98}{303}\)

\(=\dfrac{49}{303}\)

Vậy \(S=\dfrac{49}{303}\)

#\(Toru\)

chịu

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

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

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

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

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

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

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

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

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

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

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

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

Tui nghĩ là 49/303 

49/303 chắc chắn lun mình giải rùi tick nha

cô hiệu phó trường mình chữa đó.