作业帮1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)=?
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/28 09:48:55
xKN@ Ly4 X\ D5JpԁkyTEDQlwFHo;J_\ص_ `qF@K4L fPV3TqnD9v}}
wv~p ̖e,>1%>LStyΩ-eR
$ d"hҤ5{tacD16r%iaDQSExa2<\s>W|<ɩú K者{c7<н#^|t#ZDžVh5_" |
1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)=?
1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)=?
1/(1+2)+1/(1+2+3)+.+1/(1+2+3+.+100)=?
1/(1+2+……+n)
=1/[n(n+1)/2]
=2/n(n+1)
=2[1/n-1/(n+1)]
所以原式=2[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=2(1/2-1/101)
=99/404
1/(1+2+3+...+n)=1/((1+n)n/2)=2/((n+1)n)=2/n - 2/(n+1)
所以1=2/1 - 2/2
1/(1+2)=2/2 - 2/3
1/(1+2+3)=2/3 - 2/4
1/(1+2+3+4)=2/4 - 2/5
...
1/(1+2+...+n)=2/n - 2/(n+1)
将以上各式相加,
1+1/(1+2)+1/(1+2+3)+........+1/(1+2+3+....+100)
=2-2/(n+1)=2n/(n+1)=200/101
1/(1+2+……+n)
=1/[n(n+1)/2]
=2/[n(n+1)]
=2[1/n-1/(n+1)]
1/(1+2)+1/(1+2+3)+.....+1/(1+2+3+....+100)
=2[(1/2-1/3)+(1/3-1/4)+……+(1/100-1/101)]
=2(1/2-1/101)
=1-2/101
=99/101