1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=
来源：学生作业帮助网 编辑：六六作业网 时间：2024/12/29 22:11:20
1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=1*2*3/1+2*3*4/1+
1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=
1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=
1*2*3/1+2*3*4/1+3*4*5/1+``````+18*19*20/1=
1/(n-1)(n+1)n=1/2[1/n(n-1)-1/n(n+1)]
所以原式=1/2[1/(2*1)-1/(2*3)]+1/2[1/(3*2)-1/(3*4)]+.+1/2[1/(19*18)-1/(19*20)]
=1/2[1/(2*1)-1/(2*3)+1/(3*2)-1/(3*4)+.+1/(19*18)-1/(19*20)]
=1/2[1/2-1/(19*20)]
=189/760
这是我在静心思考后得出的结论,
如果能帮助到您,希望您不吝赐我一采纳~（满意回答）
如果不能请追问,我会尽全力帮您解决的~
答题不易,如果您有所不满愿意,请谅解~
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+...+1/(18*19*20)
an = 1/[n(n+1)(n+2)]
= 1/2{1/[n(n+1)] - 1/[(n+1)(n+2)]}
Sn = (1/2)( 1/2 -1/[(n+1)(n+2)] }
1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+...+1/(18*19*20)
=S18
=(1/2)( 1/2 -1/[19.20] }
= 189/760