Tô Thị Nguyệt Hà
16 tháng 8 2017 lúc 10:08
Đặt A =\(\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+....+\frac{1}{120}\) 2A \(=\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+...\frac{1}{240}\) \(=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\) \(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{15}-\frac{1}{16}\) \(=\frac{1}{4}-\frac{1}{16}\) \(=\frac{4}{16}-\frac{1}{16}\) \(=\frac{3}{16}\)