Question

[duong yen]

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+.....+\frac{2}{x\left(x-1\right)}=\frac{2007}{2009}\)

Answer 1

[Trần Đức Thắng]

ĐẶt A = \(\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)

      A = \(\frac{2}{6}+\frac{2}{12}+\frac{2}{30}\cdot\cdot\cdot+\frac{2}{x\left(x-1\right)}-\frac{2007}{2009}\)

      A = \(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+..+\frac{2}{\left(x-1\right)x}-\frac{2007}{2009}\)

      A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x-1}-\frac{1}{x}-\frac{2007}{2009}\)

A     \(=\frac{1}{2}-\frac{1}{x}-\frac{2007}{2009}\)