解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
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解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
解方程:1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
1/(x+a)(x+a+1)=1/(x+a)-1/(x+a+1)
1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+99)-1/(x+100)+1/(x+100)
=1/(x+1)=1999/2000
x=1/1999
x=1/1999
1/(x+99)(x+100)+1/(x+100)=1/(x+100)×（1/(x+99)+1）=1/(x+100)× （x+100）/(x+99)=1/(x+99)
以此类推每个的最后一项都可以消去前一项分母的后半部分
最后原式=1/(x+1)
1/(x+1)(x+2)+1/(x+2)(x+3)...+1/(x+99)(x+100)+1/(x+100)=1999/2000
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+……+1/(x+99)-1/(x+100)+1/(x+100))=1999/2000
1/(x+1)=1999/2000
(x+1)=2000/1999x=1/1999
用裂项法：1／（x+1）（x+2）=1／(x+1)－／(x+2)
每一项都裂开相加，最后为1／﹙x+1﹚=1999／2000
得出x=1／1999