求教六年级一道简算题,要有分析过程

1/(1*3)+1/(3*5)+1/(5*7)+.+1/(17*19)+1/(19*21)=1/2*2{1/(1*3)+1/(3*5)+1/(5*7)+.+1/(17*19)+1/(19*21)}2=1/2{2/(1*3)+2/(3*5)+2/(5*7)+.+1/(17*19)+2/(19*21)}=1/2*{(1-1/3)+(1/3-1/5)+(1/5-1/7)+...+(1/19-1/21)}...

1/1*3=1/1*1/3=1/2（1-1/3）
1/3*5=1/3*1/5=1/2（1/3-1/5）以此类推：
原式=1/2（1-1/3）+1/2（1/3-1/5）+1/2（1/5-1/7）+...+1/2(1/17-1/19)+1/2(1/19-1/21)
=1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/17-1/19+1/19-1/21)
=1/2(1-1/21)
=10/21

分子不变,分母两单数相乘,前一位单数比后一位单数小2 ,比前一位分母的后一位单数大2,

1/(1*3)+1/(3*5)+1/(5*7)+....+1/(17*19)+1/(19*21
因为1/(1*3）=1/2 x(1-1/3) 1/(3*5)=1/2 x(1/3 -1/5)......
原式=1/2 x { (1-1/3)+(1/3-1/5)+(1/5+1/7)......+(1/17-1/19)+(1/19-1/21)}
=1/2 x{1-1/3+1/3-1/5+1/5-1/7+1/7......+1/17-1/19+1/19-1/21}
除1和-1/21外 其余加起来为0
所以=1/2 x (1-1/21)=10/21