设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1(1)求S1 S2和Sn S1=2 S2=6 Sn=n^2+n (2) 设bn=(1/2)^an,若对一切n∈N*,均有b1+b2+b3+...bn, bk∈(1/m,m^2-6m+16/3),求m取值范围.
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![设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1(1)求S1 S2和Sn S1=2 S2=6 Sn=n^2+n (2) 设bn=(1/2)^an,若对一切n∈N*,均有b1+b2+b3+...bn, bk∈(1/m,m^2-6m+16/3),求m取值范围.](/uploads/image/z/7441239-39-9.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%B7%B2%E7%9F%A51%2FS%2B1%2FS2%2B%E2%80%A61%2FSn%3Dn%2Fn%2B1%EF%BC%881%EF%BC%89%E6%B1%82S1+S2%E5%92%8CSn+++++++++++++++++++++++++S1%3D2+S2%3D6++Sn%3Dn%5E2%2Bn++++%282%29+%E8%AE%BEbn%3D%EF%BC%881%2F2%EF%BC%89%5Ean%2C%E8%8B%A5%E5%AF%B9%E4%B8%80%E5%88%87n%E2%88%88N%2A%2C%E5%9D%87%E6%9C%89b1%2Bb2%2Bb3%2B...bn%2C++bk%E2%88%88%281%2Fm%2Cm%5E2-6m%2B16%2F3%29%2C%E6%B1%82m%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4.)
设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1(1)求S1 S2和Sn S1=2 S2=6 Sn=n^2+n (2) 设bn=(1/2)^an,若对一切n∈N*,均有b1+b2+b3+...bn, bk∈(1/m,m^2-6m+16/3),求m取值范围.
设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1
(1)求S1 S2和Sn S1=2 S2=6 Sn=n^2+n
(2) 设bn=(1/2)^an,若对一切n∈N*,均有b1+b2+b3+...bn, bk∈(1/m,m^2-6m+16/3),求m取值范围.
设数列{an}的前n项和为Sn,已知1/S+1/S2+…1/Sn=n/n+1(1)求S1 S2和Sn S1=2 S2=6 Sn=n^2+n (2) 设bn=(1/2)^an,若对一切n∈N*,均有b1+b2+b3+...bn, bk∈(1/m,m^2-6m+16/3),求m取值范围.
n等于1时
S1=2
n大于等于2时
1/S1+1/S2+…1/Sn=n/n+1 (1)
1/S1+1/S2+…1/Sn-1=n-1/n (2)
(1)-(2)得1/Sn=1/n^2+n
所以Sn=n^2+n
第二问你没写清楚:均有b1+b2+b3+...bn,bk∈(1/m,m^2-6m+16/3)?
可以求出bn的前n项和
a1=S1=2
n大于等于2时
an=Sn-Sn-1
=2n
又因为a1=2
所以an=2n
bn=(1/2)^2n
用等比数列求和的方法得:
Tn=b1+b2+b3+...bn=1/3(1-(1/4)^n)
1/4