Tìm số tự nhiên x, biết: 1/3+1/6+1/10+...+2/x.(x+1)=1999/2001

Lời giải:

`1/3+1/6+1/10+...+2/(x(x+1))=1999/2001`

`=>2/6+2/12+2/20+...+2/(x(x+1))=1999/2001`

`=>2/2.3+2/3.4+2/4.5+...+2/(x(x+1))=1999/2001`

`=>2[1/2.3+1/3.4+1/4.5+...+1/(x(x+1))]=1999/2001`

`=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=1999/4002`

`=>1/2-1/(x+1)=1999/4002`

`=>1/(x+1)=1/2001`

`=>x+1=2001`

`=>x=2000`

Vậy `x=2000`