原式=(1-1/2)+(1/2-1/3)+···+(1/2002-1/2003)+(1/2003-1/2004)+(1/2004-1/2005) =1-1/2+1/2-1/3+···+1/2002-1/2003+1/2003-1/2004+1/2004-1/2005 =1-1/2005 =2004/2005
1/(21*2)+1/(2*3)···+1/(2002*2003)+1/(2003*2004)+1/(2004*2005
1个回答
相关问题
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1.2005*2004-2004*2003+2003*2002-2002*2001+..+3*2-2*1
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1/2001*2002+1/2002*2003+1/2003*2004+1/2004*2005+1/2005=?/?
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2001/2002+2002/2003+2003/2004+2004/2005+1/2005=?
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1+2+3...+2003+2004+2005+2006+2005+2004+2003+...3+2+1
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如何解2005-2004+2003-2002+~+3-2+1
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|2003/1+2002/1|+|2004/1-2003/1|-|2004/1-2002/1|
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计算：2005×2004-2004×2003+2003×2002-2002×2001+…+3×2-2×1=______．
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计算（2+1/2003+1/2004+1/2005)*(1/2003+1/2004+1/2005+1/2006)-(2+
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计算：（2+1/2003+1/2004+1/2005）*（1/2003+1/2004+1/2005+1/2006）-（2
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初一数学在线解答1/1x2+1/2x3+.+1/2002x2003+1/2003x2004+1/2004x2005