数学题 依然分式

1/(x+1)+1/(x+1)(x+2)+.......+1/(x+2006)(x+2007)
=1/(x+1)+1/(x+1)-1/(x+2)+.......+1/(x+2006)-1/(x+2007)
=2/（x+2）-1/(x+2007)
=(x-5012)/[(x+2)(x+2007)]

有这样一个式子1/[n*(n+1)]=1/n-1/(n+1)
所以原来的式子就可以变成
1/(x+1)+1/(x+1)-1/(x+2)+.......+1/(x+2006)-1/(x+2007)=后面就自己算啦！

1/(x+1)+1/(x+1)(x+2)+.......+1/(x+2006)(x+2007)
=1/(x+1)+1/(x+1)-1/(x+2)+.......-1/(x+2006)+(x+2006)-1/(x+2007) =2/(x+1)-1/(x+2007) =（x+4013)/(x+2007)* (x+1)