计算1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+……+1/[(x+1998)(x+1999)]1/x-1/(x+1999)怎么两个除号?懂的人用分数告诉我 1————X用这样的形式.
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因为：1/(1*2)=1/2=1-1/2
1/(3*4)=1/12=1/3-1/4
.1/x(x+1)=1/x-1/(x+1)
所以,1/[x(x+1)]+1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+……+1/[(x+1998)(x+1999)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+.+1/(x+1998)-1/(x+1999)
=1/x-1/(x+1999)
=1999/[x(x+1999)]