已知数列{an}的前n项和为Sn,且满足a1=1,an+2SnS(n-1)=0(n≥2)(1)求证:{1/Sn}为等差数列 (2)设bn=Sn/2n+1 ,求数列{bn}的前n项和Tn (3)是否存在自然数m,使得对任意n属于正整数,都有Tn
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![已知数列{an}的前n项和为Sn,且满足a1=1,an+2SnS(n-1)=0(n≥2)(1)求证:{1/Sn}为等差数列 (2)设bn=Sn/2n+1 ,求数列{bn}的前n项和Tn (3)是否存在自然数m,使得对任意n属于正整数,都有Tn](/uploads/image/z/2758967-71-7.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%94%E6%BB%A1%E8%B6%B3a1%3D1%2Can%2B2SnS%EF%BC%88n-1%EF%BC%89%3D0%EF%BC%88n%E2%89%A52%EF%BC%89%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%EF%BD%9B1%2FSn%EF%BD%9D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97+%EF%BC%882%EF%BC%89%E8%AE%BEbn%3DSn%2F2n%2B1+%2C%E6%B1%82%E6%95%B0%E5%88%97%EF%BD%9Bbn%EF%BD%9D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn+%EF%BC%883%EF%BC%89%E6%98%AF%E5%90%A6%E5%AD%98%E5%9C%A8%E8%87%AA%E7%84%B6%E6%95%B0m%2C%E4%BD%BF%E5%BE%97%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E5%B1%9E%E4%BA%8E%E6%AD%A3%E6%95%B4%E6%95%B0%2C%E9%83%BD%E6%9C%89Tn)
已知数列{an}的前n项和为Sn,且满足a1=1,an+2SnS(n-1)=0(n≥2)(1)求证:{1/Sn}为等差数列 (2)设bn=Sn/2n+1 ,求数列{bn}的前n项和Tn (3)是否存在自然数m,使得对任意n属于正整数,都有Tn
已知数列{an}的前n项和为Sn,且满足a1=1,an+2SnS(n-1)=0(n≥2)
(1)求证:{1/Sn}为等差数列 (2)设bn=Sn/2n+1 ,求数列{bn}的前n项和Tn (3)是否存在自然数m,使得对任意n属于正整数,都有Tn>(m-8)/4成立?若存在,求出m的最大值;若不存在,请说明理由
已知数列{an}的前n项和为Sn,且满足a1=1,an+2SnS(n-1)=0(n≥2)(1)求证:{1/Sn}为等差数列 (2)设bn=Sn/2n+1 ,求数列{bn}的前n项和Tn (3)是否存在自然数m,使得对任意n属于正整数,都有Tn
(1).an+2SnS(n-1)=0
Sn-S(n-1)+2SnS(n-1 )=0
两边同时除以 SnS(n-1)得:
1/S(n-1)- 1/Sn +2=0
∴1/Sn - 1/S(n-1)=2 (n≥2)
当n=1时,a2=-1/2 满足上式
∴ {1/Sn}为等差数列,首项为1,公差为2.
(2).由(1)得,Sn=1/(2n-1)
∴bn=1/[(2n-1)(2n+1)]
Tn= b1+b2+……+bn
=1/(1*3)+1/(3*5)+……+1/[(2n-1)(2n+1)]
=1/2[1-1/3+1/3-1/5+……+1/(2n-1)-1/(2n+1)] (中间用拆项求和)
=1/2-1/(4n+2)(n≥1).
(3).存在.
Tn>(m-8)/4 化简得
m