设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1,设bn=(1/2)^an,数列{bn}的前n项和为T_数学解答

设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1,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

人气：368 ℃　时间：2020-07-05 00:39:39

解答

1/Sn=n/(n+1)-(n-1)/n=1/(n(n+1))1/Sn-1=(n-1)/n-(n-2)/(n-1)=1/(n(n-1))Sn=n(n+1) Sn-1=n(n-1)an=Sn-Sn-1=2n1/a1=1/S1=1/(1+1)a1=2a1也符合所以通项公式为 an=2nb1=1/4bn/bn-1=(1/2)^2n/(1/2)^(2n-2)=1/4所...