(1-1/2004)+(2-1/2004×2)+（3-1/2004×3）+......+(2004-1/2004×2004)

来源：百度知道编辑：UC知道时间：2024/06/30 15:25:59

原式＝(1-1/2004)+2*(1-1/2004)+...+2004*(1-1/2004)
=(1+2+..+2004)(2003/2004)
=2005*2004/2*2003/2004
=2005*2003/2=2008007.5

原式=(1+2+3+...+2004)-(1/2004(1+2+3+...2004))
=2009010-1002.5
=2008007.5

最简单的解法了...

(1-1/2004)+(2-1/2004×2)+（3-1/2004×3）+......+(2004-1/2004×2004)

=(1-1/2004)+2*(1-1/2004)+3*（1-1/2004）+......+2004*(1-1/2004)

=(1-1/2004)*(1+2+3+...+2004)

=2003/2004*2004*(2004+1)/2

=2003*2005/2

=4016015/2

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) 1/2+1/3+1/4+1/5******+1/2004=? (1-1/2-1/3-...-1/2003)(1/2+1/3...+1/2004)-(1-1/2-1/3-...-1/2004)(1/2+1/3...+1/2003) (1/2005-1）（1/2004-1)...(1/3-1)(1/2-1) 计算(1/2+1/3+...+1/2005)(1+1/2+1/3+...+1/2004)-(1+1/2+1/2005)(1/2+1/3+...+1/2004) 1-(1-1/2)-(1/2-1/3)-…-（1/2004-1/2005）求(1-1/2+1/3-1/4+...+1/1999)/(1/2000+1/2002+1/2004+...+1/3998) (1-1/2005)+(1-2/2005)+(1-3/2005)+..............(1-2003/2005))+(1-2004/2005) (1-1/2×2)×(1-1/3×3)×(1-1/4×4)×…×(1-1/2004×2004)= 计算:(1-1/2*2)(1-1/3*3)(1-1/4*4).....(1-1/2004*2004)