Question

tìm x thuộc N biết:

A = \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2007}{2009}\)

Answer 1

\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{2}{x\left(x+1\right)}\)

\(=2\left(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{x\left(x+1\right)}\right)=2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)\)

\(=2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{x-1}{x+1}=\dfrac{2007}{2009}\)

\(\Leftrightarrow2009x-2009=2007x+2007\)

\(\Leftrightarrow2x=4016\)

\(\Leftrightarrow x=2008\)