在数列an中,Sn是数列an前n项和,a1=1,当n≥2时,sn^2=an(Sn-1/2) (1)证明1/Sn为等差数列,并求an(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,使得对任意自然数n属于N*,都有Tn<1/4(m-8)成
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![在数列an中,Sn是数列an前n项和,a1=1,当n≥2时,sn^2=an(Sn-1/2) (1)证明1/Sn为等差数列,并求an(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,使得对任意自然数n属于N*,都有Tn<1/4(m-8)成](/uploads/image/z/693761-41-1.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2CSn%E6%98%AF%E6%95%B0%E5%88%97an%E5%89%8Dn%E9%A1%B9%E5%92%8C%2Ca1%3D1%2C%E5%BD%93n%E2%89%A52%E6%97%B6%2Csn%5E2%3Dan%28Sn-1%2F2%29+%EF%BC%881%EF%BC%89%E8%AF%81%E6%98%8E1%2FSn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%B9%B6%E6%B1%82an%EF%BC%882%EF%BC%89%E8%AE%BEbn%3DSn%2F2n%2B1%2C%E6%B1%82%E6%95%B0%E5%88%97bn%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%8F%E8%87%AA%E7%84%B6%E6%95%B0n%E5%B1%9E%E4%BA%8EN%2A%2C%E9%83%BD%E6%9C%89Tn%EF%BC%9C1%2F4%EF%BC%88m-8%EF%BC%89%E6%88%90)
在数列an中,Sn是数列an前n项和,a1=1,当n≥2时,sn^2=an(Sn-1/2) (1)证明1/Sn为等差数列,并求an(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,使得对任意自然数n属于N*,都有Tn<1/4(m-8)成
在数列an中,Sn是数列an前n项和,a1=1,当n≥2时,sn^2=an(Sn-1/2) (1)证明1/Sn为等差数列,并求an
(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,使得对任意自然数n属于N*,都有Tn<1/4(m-8)成立?若存在,求出m的最小值,若不存在,请说明理由
在数列an中,Sn是数列an前n项和,a1=1,当n≥2时,sn^2=an(Sn-1/2) (1)证明1/Sn为等差数列,并求an(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,使得对任意自然数n属于N*,都有Tn<1/4(m-8)成
(Sn)²=[Sn-S(n-1)](Sn-1/2)
(Sn)²=(Sn)²-Sn/2-SnS(n-1)+S(n-1)/2
Sn+2SnS(n-1)-S(n-1)=0
S(n-1)-Sn=2SnS(n-1)
两边除以SnS(n-1)
1/Sn-1/S(n-1)=2
1/Sn等差,d=2
S1=a1=1
1/Sn=1/S1+2(n-1)=2n-1
Sn=1/(2n-1)
bn=1//[(2n-1)(2n+1)]
=1/2*2[(2n-1)(2n+1)]
=1/2*[(2n+1)-(2n+1)]/[(2n-1)(2n+1)]
=1/2*{(2n+1)/[(2n-1)(2n+1)]-(2n+1)/[(2n-1)(2n+1)]}
=1/2*[1/[(2n-1)-1/(2n+1)]
所以Tn=1/2*(1-1/3+1/3-1/5+1/5-1/7+……+1/[(2n-1)-1/(2n+1)]
=1/2*(1-1/(2n+1)]
=n/(2n+1)
Tn=1/(2+1/n) 随n增加而递增.
1/3=T1