1/1×3+1/3×5+1/5×7+……+1/2003×2005

来源：百度知道编辑：UC知道时间：2024/05/25 05:08:42

解：原式=（1/1-1/3+1/3-1/5+1/5-1/7+……+1/2003-1/2005）×1/2
=（1-1/2005）×1/2
=2004/2005×1/2
=1002/2005

解:原式=1/2(1-1/3)+1/2(1/3 -1/5)+...+1/2(1/2003 -1/2005)
=1/2[(1/1-1/3)+(1/3-1/5)+...(1/2003-1/2005)
=1/2(1-1/2005)
=1002/2005

注意满足：1/n（n+2）=1/2(1/n -1/n+2)

原式=1/2[(1/1-1/3)+(1/3-1/5)+...(1/2003-1/2005)
=1/2(1-1/2005)
=1002/2005

(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1) (1-1/100)(1-1/99)(1-1/98)......(1-1/3 1+1/2+1/3+.....+1/n 1+1/2+1/3+...+1/100 1+1/3+1/6+........+1/55 1-1/2+1/3-.....-1/10 (1+1/2+1/3+1/4)× (1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000) 3,1/3,1/3,1/,1/27,1/27( ) 1+1/2+1+1/3+1+1/4+......+1/100=?