Question

Tìm x, biết

1/3+1/6+1/10+...+2/x(x+1)=2005/2007

Answer 1

Ta có :\(\frac{1}{3}+\frac{1}{6}+..+\frac{2}{x\left(x+1\right)}=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)

          = 2 x \(\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...\frac{1}{x\left(x+1\right)}\right)=2\times\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)\)

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

          = 2 x (\(\frac{1}{2}-\frac{1}{x+1}\)

       Khi đó chỉ cần giải  2 x\(\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2005}{2007}\)