Question

Tìm x biết : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+..,+\frac{2}{x\left(x+1\right)}=\frac{2004}{2005}\)

Answer 1

Lời giải:

$\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x(x+1)}=\frac{2004}{2005}$

$2(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)})=\frac{2004}{2005}$

$\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x(x+1)}= \frac{1002}{2005}$

$\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x(x+1)}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{2}-\frac{1}{x+1}=\frac{1002}{2005}$

$\frac{1}{x+1}=\frac{1}{2}-\frac{1002}{2005}=\frac{1}{4010}$

$\Rightarrow x+1=4010$

$\Rightarrow x=4009$