(1/2) [ 1/x - 1/(x+2) + 1/(x+2) - 1/(x+4) + 1/(x+4) - 1/(x+6) + ... +1/(x+2004) - 1/(x+2006)]
=(1/2) [ 1/x - 1/(x+2006)]
=1003/[x(x+2006)]
解析:根据原题给出的解法,你可以做出大胆的推理,即1/x*1/(x+2)能否拆成题干所给的形式。所以你就可以想1/x和1/(x+2)能不能拆成“?-?”的形式。然后你就会发现可以把这两个式子相减再除以2和原式相等。接着依此类推。再结合题干的计算方法,发现除了第一项和最后一项外,其余的都能合并,化0。所以只需把1/2提出来,首相和末相相减。即可得出答案。
原式=(1/2) [ 1/x - 1/(x+2) + 1/(x+2) - 1/(x+4) + 1/(x+4) - 1/(x+6) + ... +1/(x+2004) - 1/(x+2006)]
=(1/2) [ 1/x - 1/(x+2006)]
=1003/[x(x+2006)]
0.5*[1/x-1/(x+2)+1/(x+2)-1/(x+4).....+1/(x+2004)-1/(x+2006)]=1003/x(x+2006)