1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)
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1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+..
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+100)-1/(x+101)
=1/(x+1)-1/(x+101)
=(x+101-x-1)/(x+1)(x+101)
=100/(x+1)(x+101)
答：分式的裂项
1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+100)(x+101)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+4)+....+1/(x+100)-1/(x+101)
=1/(x+1)-1/(x+101) (中间各项正负抵消，剩余第一项和最后一项）
=100/[(x+1)(x+101)]
因为1/(x+1)(x+2)=1/x+1 - 1/x+2
同理1/(x+100)x+(101)=1/x+100 - 1/x+101.....
所以原式=1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 +...+ 1/x+100 - 1/x+101 = 1/x+1 - 1/x+101 = 100/(x+1)(x+101)
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