如题所述
1+2+3+。。。+n=n(n+1)/2
1/(1+2+3+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
原式=2[1-1/2+1/2-1/3++1/3-1/4+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)=2x2013/（2013+1）
=4026/2014=2013/1007追问
1/(1+2+3+...+n)=2/[n(n+1)]=2[1/n-1/(n+1)]
原式=2[1-1/2+1/2-1/3++1/3-1/4+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)=2x2013/（2013+1）
=4026/2014=2013/1007追问
3/5+2/5/3/5+2/5
追答3/5+2/5/3/5+2/5
=1+2/3
=5/3
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第1个回答 2013-06-07
sum_{k=2}^{2013}1/(1+2+...+k)
=sum_{k=2}^{2013}2/k/(k+1)
=2 sum_{k=2}^{2013}(1/k-1/(k+1))
=2(1/2-1/2014)
=1-1/1007
=1006/1007.追问
=sum_{k=2}^{2013}2/k/(k+1)
=2 sum_{k=2}^{2013}(1/k-1/(k+1))
=2(1/2-1/2014)
=1-1/1007
=1006/1007.追问
3/5+2/5/3/5+2/5
追答1+2/75=77/75.
