Question

 1/4+ 1/12 + 1/36+ 1/108+....+ 1/2010 + 1/6030

Answer 1

Đặt \(S=\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{2010}+\frac{1}{6030}.\)

        \(\Rightarrow3S=\frac{3}{4}+\frac{3}{12}+\frac{3}{36}+...+\frac{3}{2010}+\frac{3}{6030}\)

                   \(=\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{670}+\frac{1}{2010}\)

  \(\Rightarrow3S-S=2S=\left(\frac{3}{4}+\frac{1}{4}+\frac{1}{12}+...+\frac{1}{670}+\frac{1}{2010}\right)-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{36}+...+\frac{1}{2010}+\frac{1}{6030}\right)\)

\(2S=\frac{3}{4}-\frac{1}{6030}\)

\(\Rightarrow S=\frac{\frac{3}{4}-\frac{1}{6030}}{2}\)