Axiao_zz
幼苗
共回答了17个问题采纳率:100% 举报
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...100)=?
1/(1+2+3+...n)=1/[n(1+n)/2]=2/n(n+1)=2(1/n-1/(n+1))
所以原式=2(1-1/2)+2(1/2-1/3)+2(1/3-1/4)+...+2(1/100-1/101)
=2(1-1/2+1/2-1/3+1/3-1/4+...+1/100-1/101)
=2(1-1/101)
=200/101
1年前
10