Question

Tìm x thuộc N biết: 1/3+1/6+1/10+...+2/x(x+4)=2001/2003

Answer 1

Sửa đề: \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2001}{2003}\)

\(\Leftrightarrow2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2001}{2003}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2001}{4006}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2003}\)

=>x=2002