x(x+1)分之一加(x+1)(x+2)分之一…一直加到(x+2005)(x+2006)分之一 等于多少?第一个给出正确答案的在追加100分

1/(x+n)(x+n+1)=1/(x+n)-1/(x+n+1)
裂项相消
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-...-1/(x+2005)+1/(x+2005)
-1/(x+2006)
=1/x-1/(x+2006)
=2006/x(x+2006)

2006/x(x+2006)
用列项求和

我们知道1/[x(x+1)]=1/x-1/(x+1)
那么原式=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+)]+[1/(x+2)-1/(x+3)]+....+[1/(x+2005)-1/(x+2006)]
=1/x-1/(x+2006)
=2006/[x(x+2006)]

答案是1/X - 1/(X+2006)
现在我来给你写做题的步骤
因为1/X(X+1)=1/X-1/(X+1)
1/(X+1)*(X+2)=1/(X+1)-1/(X+2)
1/(X+2)*(X+3)=1/(X+2)-1/(X+3)
.....
.......
........
1/(X+2005)*(X+2006)=1/(X+2005)-1/(X+2006)
所以原式=1/X - 1/(X+2006)

1/[x(x+1)]+1/[(x+1)(x+2)]+......+1/[(x+2005)(x+2006)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+......+1/(x+2005)-1/(x+2006)
=1/x-1/(x+2006)

把X(X+1)分之一变形为X分之一减X+1分之一 把(X+1)(X+2)分之一变化为X+1分之一减X+2分之一............都打开你就知道怎么做了

1/x-1/(x+2006)

1/x(x+1)+1/(x+1)(x+2)+…+1/(x+2005)(x+2006)
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+……+[1/(x+2005)-1/(x+2006)]
=1/x-1/(x+1)+1/(x+1)-1/(x+2)+……+1/(x+2005)-1/(x+2006)
=1/x-1/(x+2006)
=2006/x(x+2006)

X(X+1)=1/X-1/(X+1)
(X+1)(X+2)=1/(X+1)-1/(X+2)
同理：各项都可以分解成这样
最后一项=1/(X+2005)-1/(X+2006）
相加得1/X-1/(X+2006)

运用裂项消去法x/[x(x+1)]=[1/x]-[1/(x+1)]~~~~~~~同理最后一项=[1/(x-2005)]-[1/(x-2006)]~~~~~中间的全消去啦然后把第一项与最后一项相加,自己试下

1/[x(x+1)]=1/x-1/(x+1)
1/[(x+1)(x+2)]=1/(x+1)-1/(x+2)
原式=1/x-1/(x+1)+1/(x+1)-1/(x+2)+....+1/(x+2005)-(x+2006)
=1/x-1/(x+2006)
=2006/x(x+2006)

原式=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+)]+[1/(x+2)-1/x+3)] +....+[1/(x+2005)-1/(x+2006)]
=1/x-1/(x+2006)

x(x+1)分之一加(x+1)(x+2)分之一…一直加到(x+2005)(x+2006)分之一
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+……+[1/(x+2005)-1/(x+2006)]
=1/x-1/(x+2006)。