已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1(n∈N*),等差数列{bn}中,bn>0(n∈N*)且
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已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1(n∈N*),等差数列{bn}中,bn>0(n∈N*)且b1+b2+b3=15,又a1+b1、a2+b2、a3+b3成等比数列.求数列{an}、{bn}的通项公式.
![已知数列{an}的前n项和为Sn,a1=1,an+1=2Sn+1(n∈N*),等差数列{bn}中,bn>0(n∈N*)且](/uploads/image/z/9823654-46-4.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%EF%BC%8Ca1%3D1%EF%BC%8Can%2B1%3D2Sn%2B1%EF%BC%88n%E2%88%88N%2A%EF%BC%89%EF%BC%8C%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%7Bbn%7D%E4%B8%AD%EF%BC%8Cbn%EF%BC%9E0%EF%BC%88n%E2%88%88N%2A%EF%BC%89%E4%B8%94)
(1)当n≥2时,由an+1=2Sn+1得an=2Sn-1+1,两式相减得
an+1-an=2Sn-2Sn-1=2an,整理得
an+1
an=3,
a2=2S1+1=3,∴
a2
a1=3满足上式.
∴{an}是以1为首项,3为公比的等比数列.
∴an=3n-1
(2)由条件知:b2=5,故(1+b1)(9+b3)=64
即(6-d)(14+d)=64,解得d=2或d=-10(舍),故b1=3
∴bn=b1+(n-1)d=2n+1
an+1-an=2Sn-2Sn-1=2an,整理得
an+1
an=3,
a2=2S1+1=3,∴
a2
a1=3满足上式.
∴{an}是以1为首项,3为公比的等比数列.
∴an=3n-1
(2)由条件知:b2=5,故(1+b1)(9+b3)=64
即(6-d)(14+d)=64,解得d=2或d=-10(舍),故b1=3
∴bn=b1+(n-1)d=2n+1
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