Đặt A=1/3+1/6+1/10+...+1/(x(x+1))
=> A=2/6+2/12+2/20+...+2/2(x(x+1))
=> 1/2A=1/(2.3)+1/(3.4)+...+1/2(x(x+1))
=> 1/2A=1/2-1/3+1/3-1/4+...+1/x-1/(x+1)
=> 1/2A=1/2-1/(x+1). Vì A=2000/2002=1000/1001=> 1/2A=500/1001
=> 1/2-1/(x+1)=500/1001
=> 1/(x+1)=1001/2002-1000/2002
=> 1/(x+1)=1/2002
=> x+1=2002
=> x=2001