已知数列{an}的前n项的和为Sn,且1/S1+1/S2+ +1/Sn=n/(n+1) (1)求S1,S2及Sn.(2)设bn=(1/2)^an,数列{bn}的前n项和为Tn,若对一切n∈N*均有Tn∈(1/m ,m^2-6m+16/3 ),求实数m的取值范围.
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![已知数列{an}的前n项的和为Sn,且1/S1+1/S2+ +1/Sn=n/(n+1) (1)求S1,S2及Sn.(2)设bn=(1/2)^an,数列{bn}的前n项和为Tn,若对一切n∈N*均有Tn∈(1/m ,m^2-6m+16/3 ),求实数m的取值范围.](/uploads/image/z/4538211-51-1.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E7%9A%84%E5%92%8C%E4%B8%BASn%2C%E4%B8%941%2FS1%2B1%2FS2%2B+%2B1%2FSn%3Dn%2F%28n%2B1%29+%EF%BC%881%EF%BC%89%E6%B1%82S1%2CS2%E5%8F%8ASn.%EF%BC%882%EF%BC%89%E8%AE%BEbn%3D%281%2F2%29%5Ean%2C%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%8B%A5%E5%AF%B9%E4%B8%80%E5%88%87n%E2%88%88N%2A%E5%9D%87%E6%9C%89Tn%E2%88%88%EF%BC%881%2Fm+%2Cm%5E2-6m%2B16%2F3+%29%2C%E6%B1%82%E5%AE%9E%E6%95%B0m%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
已知数列{an}的前n项的和为Sn,且1/S1+1/S2+ +1/Sn=n/(n+1) (1)求S1,S2及Sn.(2)设bn=(1/2)^an,数列{bn}的前n项和为Tn,若对一切n∈N*均有Tn∈(1/m ,m^2-6m+16/3 ),求实数m的取值范围.
已知数列{an}的前n项的和为Sn,且1/S1+1/S2+ +1/Sn=n/(n+1) (1)求S1,S2及Sn.
(2)设bn=(1/2)^an,数列{bn}的前n项和为Tn,若对一切n∈N*均有Tn∈(1/m ,m^2-6m+16/3 ),求实数m的取值范围.
已知数列{an}的前n项的和为Sn,且1/S1+1/S2+ +1/Sn=n/(n+1) (1)求S1,S2及Sn.(2)设bn=(1/2)^an,数列{bn}的前n项和为Tn,若对一切n∈N*均有Tn∈(1/m ,m^2-6m+16/3 ),求实数m的取值范围.
1)令n=1,得1/S1=1/(1+1)=1/2,所以S1=2;
令n=2,得1/S1+1/S2=2/(2+1),且S1=2,得S2=6
因为,1/S1+1/S2+……+1/Sn=n/(n+1)
所以,1/S1+1/S2+……+1/S(n-1)=(n-1)/(n-1+1)=(n-1)/n
两式相减,得1/Sn=n/(n+1)-(n-1)/n=1/(n*(n+1))
所以,Sn=n*(n+1)
因为,S(n-1)=(n-1)*(n-1+1)=(n-1)*n,
所以,an=Sn-S(n-1)=n*(n+1)-(n-1)*n=2n
2)bn=(1/2)^an=(1/2)^(2n)=(1/4)^n为等比数列,公比q为1/4小于1,且b1=1/4,
所以,Tn最小在n=1时取得,为1/4,最大在n为无穷大时取得,为b1/(1-q)=1/3
要使条件成立,则应满足,1/m4.
另外,m^2-6m+16/3>1/3,所以m^2-6m+5>0,解得,m>5,综上,m应该大于5.
仅供参考.