{an}为等差数列,其前n项和为Sn,a3=1/2.S3=6.
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{an}为等差数列,其前n项和为Sn,a3=1/2.S3=6.
求1/S1+1/S2+1/S3+...+1/Sn
求1/S1+1/S2+1/S3+...+1/Sn
有S3=6得a2=S3/3=2
因为a3=1/2
所以d=a3-a2=-3/2
a1=2+3/2=7/2
Sn=na1-3n(n-1)/4=7n/2-(3n^2-3n)/4=n(17-3n)/4
1/Sn=4/(n(17-3n))=12/(3n(17-3n))=12/17*(1/3n+1/(17-3n))
1/S1+1/S2+1/S3+...+1/Sn=12/17*(1/3n的和+1/(17-3n)的和)
因为a3=1/2
所以d=a3-a2=-3/2
a1=2+3/2=7/2
Sn=na1-3n(n-1)/4=7n/2-(3n^2-3n)/4=n(17-3n)/4
1/Sn=4/(n(17-3n))=12/(3n(17-3n))=12/17*(1/3n+1/(17-3n))
1/S1+1/S2+1/S3+...+1/Sn=12/17*(1/3n的和+1/(17-3n)的和)
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