高手赐教，数学题，主要题型应该是化简求值

=(a+1-a)/a(a+1)+((a+2)-(a+1))/(a+1)(a+2)+((a+3)-(a+2))+......+((a+2004)-
(a+2003))/(a+2003)(a+2004)
=1/a-1/(a+1)+)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+......+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=2003/a(a+2004)

注意到：1/a*(a+1)]=1/a-1/(a+1)
1/(a+1)*(a+2)=1/(a+1)-1/(a+2)
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1/(a+2003)*(a+2004)=1/(a+2003)-1/(a+2004)
原式=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-......+1/(a+2003)-1/(a+2004)
=1/a-1/(a+2004)
=2003/a(a+2004)
很高兴为你解决问题！

[1/a*(a+1)]+[1/(a+1)*(a+2)]+[1/(a+2)*(a+3)]+....[1/(a+2003)*(a+2004)]
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+....-1/(a+2004)
=1/a-1/(a+2004)
=2004/[(a^2+2004a)

这是典型的裂项相消类的题目。
1/n*(n+1)=1/n-1/(n+1)

[1/a*(a+1)]+[1/(a+1)*(a+2)]+[1/(a+2)*(a+3)]+....[1/(a+2003)*(a+2004)]
=1/a-1/(a+2004)
=2003/a(a+2004)

括号弄错了
应该是这样吧1/[a*(a+1)]+1/[(a+1)*(a+2)]+1/[(a+2)