等差数列an,前n项和为sn,a3=6,s3=12
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等差数列an,前n项和为sn,a3=6,s3=12
求证:1/s1+1/s2+...+1/sn
求证:1/s1+1/s2+...+1/sn
易知:d=2,sn=1/n(n+1)= 1/n - 1/(n+1)
所以:1/s1+1/s2+...+1/sn=1/1x2 + 1/2x3+.1/nx(n+1)
=1/1-1/2+1/2-1/3+.+1/n-1/(n+1)=1-1/(n+1)
所以:1/s1+1/s2+...+1/sn=1/1x2 + 1/2x3+.1/nx(n+1)
=1/1-1/2+1/2-1/3+.+1/n-1/(n+1)=1-1/(n+1)
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