Question

tìm x \(\in\)N

\(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)+\(\dfrac{1}{10}\)+....+\(\dfrac{2}{x\left(x+1\right)}\)=\(\dfrac{2000}{2002}\)

Answer 1

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2000}{2002}\)

=> \(2.\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2000}{2002}\)

=> 2.\(\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2000}{2002}\)

=> 2.\(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2000}{2002}\)

=> 2.\(\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2000}{2002}\)

=> 1-\(\dfrac{2}{x+1}-\dfrac{2000}{2002}=0\)

=> \(1-\dfrac{2000}{2002}=\dfrac{2}{x+1}\)

=> \(\dfrac{2}{2002}=\dfrac{2}{x+1}\)

=> x+1=2002

=> x=2002-1=2001