数学题（加分的）

裂项法
1/1*2=1-1/2
1/2*3=1/2-1/3
.....
1/2001*2002=1/2001-1/2002
原式
=1-1/2+1/2-1/3+1/3-1/4+.....+1/99-1/100
=1-(1/2-1/2)-(1/3-1/3)-(1/3-1/3)-.....-(1/2001-1/2001)-1/2002
=1-0-0-0-....-0-1/2002
=2001/2002

解：对原题进行化简，这种分数计算都有裂项公式的，
公式：1/1x2+1/2x3+1/3x4+1/4x5+....+1/n(n+1)
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+.....+(1/n-1/n+1)
根据公式有：
1/1x2+1/2x3+1/3x4+1/4x5+....+1/2000x2001+1/2001x2002
=(1-1/2)+(1/2-1/3)+(1/3-1/4)+(1/4-1/5)+.....+(1/2000-1/2001)+(1/2001-1/2002)
=1-1/2002
=2001/2002