已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2(n-1次)an设bn=an/an+1求bn的前n项和Tn
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![已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2(n-1次)an设bn=an/an+1求bn的前n项和Tn](/uploads/image/z/2117879-71-9.jpg?t=%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%E6%BB%A1%E8%B6%B3an%E2%89%A00%2CanSn%2B1-an%2B1Sn%3D%282%E7%9A%84n-1%E6%AC%A1%29an%2B1an%2C%E6%B1%82Sn%3D2%EF%BC%88n-1%E6%AC%A1%EF%BC%89an%E8%AE%BEbn%3Dan%2Fan%2B1%E6%B1%82bn%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn)
已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2(n-1次)an设bn=an/an+1求bn的前n项和Tn
已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2(n-1次)an设bn=an/an+1
求bn的前n项和Tn
已知数列{an}的前n项和为Sn,满足an≠0,anSn+1-an+1Sn=(2的n-1次)an+1an,求Sn=2(n-1次)an设bn=an/an+1求bn的前n项和Tn
(1)anS(n+1)-a(n+1)Sn=2^(n-1)a(n+1)an
两边同除a(n+1)an得:
S(n+1)/a(n+1)-Sn/an=2^(n-1)
设cn=Sn/an
∴c(n+1)=cn+2^(n-1)
∴c(n+1)-2^n=cn-2^(n-1)=...=c1-2^0=S1/a1-1=0
∴cn=2^(n-1)
∴Sn/an=2^(n-1)
∴Sn=2^(n-1)an
(2)Sn=2^(n-1)an
S(n+1)=2^na(n+1)
∴a(n+1)=S(n+1)-Sn=2^na(n+1)-2^(n-1)an
∴2^(n-1)an=(2^n-1)a(n+1)
∴bn=an/a(n+1)=(2^n-1)/2^(n-1)=2-1/2^(n-1)
∴Tn=2n-1(1-1/2^n)/(1-1/2)=2n-[2-(1/2)^(n-1)]
这是我在静心思考后得出的结论,
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