已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设b
已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设bn=log3an,求Tn=b1b2-b2b3+...
已知等比数列{an}的前n项和为Sn,a1=3,且3S1,2S2,S3成等差数列.(1)求数列{an}的通项公式;(2)设bn=log3an,求Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1.
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(Ⅰ)∵3S1,2S2,S3成等差数列,∴4S2=3S1+S3,
∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,
∴an=a1qn-1=3n.…(6分)
(Ⅱ)由(Ⅰ)知,bn=log3an=log33n=n,
∵b2n-1b2n-b2nb2n+1=(2n-1)?2n-2n(2n+1)=-4n,
∴Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1
=-4(1+2+3+…+n)=-4×
=-2n2-2n.…(12分)
∴4(a1+a2)=3a1+(a1+a2+a3),即a3=3a2,∴公比q=3,
∴an=a1qn-1=3n.…(6分)
(Ⅱ)由(Ⅰ)知,bn=log3an=log33n=n,
∵b2n-1b2n-b2nb2n+1=(2n-1)?2n-2n(2n+1)=-4n,
∴Tn=b1b2-b2b3+b3b4-b4b5+…+b2n-1b2n-b2nb2n+1
=-4(1+2+3+…+n)=-4×
n(n+1) |
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