1⼀2*4+1⼀4*6+1⼀6*8+····+1⼀2006*2008的过程！！~我不懂啊！！~急！！~

解：an=1/2n（2n+2）=1/4×[1/n（n+1）]=1/4×[1/n-1/（n+1）]
1/2*4=1/4×（1-1/2)
1/4*6=1/4×（1/2-1/3)
1/6*8=1/4×（1/3-1/4)
……
1/2006*2008=1/4×（1/2003-1/2004)
∴1/2*4+1/4*6+1/6*8+····+1/2006*2008
=1/4×（1-1/2)+1/4×（1/2-1/3)+1/4×（1/3-1/4)+……+1/4×（1/2003-1/2004)
=1/4×（1-1/2+1/2-1/3+1/3-1/4+……+1/2003-1/2004)
=1/4×（1-1/2004)
=1/4×2003/2004
=2003/8016

1/2*4+1/4*6+1/6*8+····+1/2006*2008
=1/2(1/2-1/4)+1/2(1/4-1/6)+1/2(1/6-1/8)+...+1/2(1/2006-1/2008)
=1/2(1/2-1/2008)
=1003/4016

原式=1/2*(1/2-1/4+1/4-1/6+1/6-1/8+.......+1/2006-1/2008)
=1/2*(1/2-1/2008)
=1/2*1003/2008
=1003/4016