1/(a-1)+1/(a-1)(a-2)+1/(a-2)(a-3)+1/(a-3)(a-4）……1/(a-2004)(a-2005)
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1=(a-1)-(a-2)
所以1/(a-1)(a-2)=[(a-1)-(a-2)]/(a-1)(a-2)=(a-1)/(a-1)(a-2)-(a-2)/(a-1)(a-2)=1/(a-2)-1/(a-1)
同理,1=(a-2)-(a-3),得到
1/(a-2)(a-3)=1/(a-3)/1/(a-2)
后面以此类推
所以1/(a-1)+1/(a-1)(a-2)+1/(a-2)(a-3)+1/(a-3)(a-4）……1/(a-2004)(a-2005)
=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)/1/(a-2)+……+1/(a-2005)-1/(a-2004)
=1/(a-2005)
所以1/(a-1)(a-2)=[(a-1)-(a-2)]/(a-1)(a-2)=(a-1)/(a-1)(a-2)-(a-2)/(a-1)(a-2)=1/(a-2)-1/(a-1)
同理,1=(a-2)-(a-3),得到
1/(a-2)(a-3)=1/(a-3)/1/(a-2)
后面以此类推
所以1/(a-1)+1/(a-1)(a-2)+1/(a-2)(a-3)+1/(a-3)(a-4）……1/(a-2004)(a-2005)
=1/(a-1)+1/(a-2)-1/(a-1)+1/(a-3)/1/(a-2)+……+1/(a-2005)-1/(a-2004)
=1/(a-2005)
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