flecture
幼苗
共回答了23个问题采纳率:87% 举报
分母等差数列求和:
1+2+3+……+n
=n(n+1)/2
1/[n(n+1)/2]=2/n(n+1)
而1/n(n+1)=1/n-1/(n+1)
1+ 1/(1+2) + 1/(1+2+3) + 1/(1+2+3+4)…… 1/(1+2+3+.100)
=1+2/2×3+2/3×4+2/4×5+……+2/100×101
=1+2×(1/2-1/3+1/3-1/4+1/4-1/5+……+1/99-1/100)
=1+2×(1/2-1/101)
=1+1-2/101
=200/101
1年前
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