(1/2)已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnS(n-1)=0 (n>=2,n属于N) (1)求an和Sn
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![(1/2)已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnS(n-1)=0 (n>=2,n属于N) (1)求an和Sn](/uploads/image/z/4019987-11-7.jpg?t=%281%2F2%29%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%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%2F2%2Can%2B2SnS%28n-1%29%3D0+%28n%3E%3D2%2Cn%E5%B1%9E%E4%BA%8EN%29+%281%29%E6%B1%82an%E5%92%8CSn)
(1/2)已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnS(n-1)=0 (n>=2,n属于N) (1)求an和Sn
(1/2)已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnS(n-1)=0 (n>=2,n属于N) (1)求an和Sn
(1/2)已知数列{an}的前n项和为Sn,且满足a1=1/2,an+2SnS(n-1)=0 (n>=2,n属于N) (1)求an和Sn
an+2SnS(n-1)=0 an=sn-s(n-1)
所以Sn-S(n-1)+2SnS(n-1)=0 除以SnS(n-1)
1/Sn-1/S(n-1) =2
1/Sn等差
1/Sn=1/S1+2(n-1)=2n
Sn=1/2n
an=Sn-Sn-1=1/2n-1/2(n-1)=-1/2n(n-1) n>=2
an=1/2 n=1
=-1/2n(n-1) n>=2
an+2SnS(n-1)=0,考虑到an=Sn-S(n-1),得:
Sn-S(n-1)+2SnS(n-1)=0
S(n-1)-Sn=2SnS(n-1) 【两边除以SnS(n-1)】
1/[Sn]-1/[S(n-1)]=2=常数
则数列{1/Sn}是以1/S1=2为首项、以d=2为公差的等差数列,得:
1/Sn=2n
Sn=1/(2n)...
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an+2SnS(n-1)=0,考虑到an=Sn-S(n-1),得:
Sn-S(n-1)+2SnS(n-1)=0
S(n-1)-Sn=2SnS(n-1) 【两边除以SnS(n-1)】
1/[Sn]-1/[S(n-1)]=2=常数
则数列{1/Sn}是以1/S1=2为首项、以d=2为公差的等差数列,得:
1/Sn=2n
Sn=1/(2n)
则:an= { 1/2 (n=1)
{ Sn-S(n-1)=1/[2n(1-n)] (n≥2)
【本题必须分段】
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