Question

[Alan Walker]

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)}=\frac{1999}{2001}\)

Tìm x.

Answer 1

[ST]

\(\Rightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{1}{2}\cdot\frac{1999}{2001}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1999}{4002}\)

\(\Rightarrow\frac{1}{x+1}=\frac{1}{2001}\)

=> x + 1 = 2001

=> x = 2000

Answer 2

[Lê Hải Anh]

x=2000