1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) + ...+1/ (x+2005)(x+2006) = 1/2x + 4012
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1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+...+1/(x+2005)(x+2006)=1/2x+40121/(x+1)(x+2)+1/(x+2)(x+3)+1/(
1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) + ...+1/ (x+2005)(x+2006) = 1/2x + 4012
1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) + ...+1/ (x+2005)(x+2006) = 1/2x + 4012
1/(x+1)(x+2) + 1/(x+2)(x+3) + 1/(x+3)(x+4) + ...+1/ (x+2005)(x+2006) = 1/2x + 4012
由于1/(x+i)(x+i+1)可以裂开成1/(x+i)-1/(x+i+1)
所以每一项被裂开成两部分 每项的后一部分和下一项的前一部分地抵消 只有第一项的前一部分和最后一项的后一部分没有被抵消
故原式=1/(x+1)-1/(x+2006)
根据规律1/(x+1)(x+2) = 1/(x+1) -1/(x+2)就可以了!
最后会剩下 1/(x+1) -1/(x+2006) = 1/2+4012 ,接下来的自己动手,丰衣足食。
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