|a-1| ≥ 0,|ab-2| ≥ 0
|a-1| + |ab-2| = 0
因此 a-1 = 0,ab - 2 = 0解得 a = 1,b = 2.原式化为:
1/(1x2) + 1/(2x3) + 1/(3x4) + ...+ 1/(2001x2002)
=1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/1999 - 1/2000 + 1/2000 - 1/2001
=1 - 1/2001
=2000/2001
|a-1| ≥ 0,|ab-2| ≥ 0
|a-1| + |ab-2| = 0
因此 a-1 = 0,ab - 2 = 0解得 a = 1,b = 2.原式化为:
1/(1x2) + 1/(2x3) + 1/(3x4) + ...+ 1/(2001x2002)
=1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ...+ 1/1999 - 1/2000 + 1/2000 - 1/2001
=1 - 1/2001
=2000/2001