1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=

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1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=1/(2x4)+1/(4x6)+

1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=
1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=

1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=
1/(2x4)+1/(4x6)+1/(6x8)+...+1/(2006x2008)=1/2[(1/2-1/4)+(1/4-1/8)+…+(1/2006-1/2008)]=1/2(1/2-1/2008)=1003/4016

1/(2×4)+1/(4×6)+1/(6×8)+…+1/(2006×2008)
=1/2×(1/2-1/4+1/4-1/6+1/6-1/8+…1/2006-1/2008)
=1/2×(1/2-1/2008)
=1003/4016

原式=1/2(1/(1*2)+1/(2*3)。。。+1/(1003*1004))
=1/2(1/1-1/2+1/2-1/3+1/3.。。。+1/1003-1/1004)
=1/2(1-1/1004)
=1/2*1003/1004
=1003/2008

原式=1/2(1/1*2+1/2*3+1/3*4""""+1/1003*1004)=1/2(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5"""-1/1003+1/1003-1/1004)=1/2(1-1/1004)=1/2 *1003/1004 =1003/2008

每一项现提出个1/2,得到1/2(1-1/2+1/2-1/3+1/3-1/4.........+1/1003+1/1004),最后得到1/2(1-1/1004)=3/2008,典型的列项思想

(1/(2x4))+(1/(4x6)+(1/(6x8))+...+(1/(2006x2008))
=1/(2x1x2x2)+1/(2x2x2x3)+1/(2x3x2x4)+···+1/(2x1003x2x1004)
=1/4【1/(1x2)+1/(2x3)+1/(3x4)+···+1/(1003x1004)】
=(1/4)x(1003/1004)
=1003/4016