计算1/（x-1)+1/（（x-1)(x-2))+1/((x-2)(x-3))+......+1/((x-99)(x-100))

1/[(x-1)(x-2)]=1/(x-2)-1/(x-1).
依此类推，原式＝1/(x-1)+1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)......+1/(x-100)-1/(x-99)=1/(x-100)

1/（x-1)+1/（（x-1)(x-2))
=(x-2)/((x-1)(x-2)))+1/（（x-1)(x-2))
=(x-2+1)/(（x-1)(x-2))
=(x-1)/((x-1)(x-2))
=1/(x-2)
依次类推
因为1/（x-1)+1/（（x-1)(x-2))=1/(x-2)
可以用相同的方法推出
1/(x-2)+1/((x-2)(x-3))=1/(x-3)
1/(x-3)+1/((x-3)(x-4))=1/(x-4)
......直到
1/(x-99)+1/((x-99)(x-100))=1/(x-100)
所以答案是1/(x-100)

答案是1/(x-100)

汗