数列{an}满足a1=2,a2=5,an+2=3an+1-2an.
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数列{an}满足a1=2,a2=5,an+2=3an+1-2an.
(1)求证:数列{an+1-an}是等比数列;
(2)求数列{an}的通项公式;
(3)求数列{an}的前n项和Sn.
(1)求证:数列{an+1-an}是等比数列;
(2)求数列{an}的通项公式;
(3)求数列{an}的前n项和Sn.
(1)由题意知:an+2-an+1=2(an+1-an).
∴
an+2−an+1
an+1−an=2,故数列{an+1−an}是等比数列(4分).
(2)由(1)知数列{an+1-an}以是a2-a1=3为首项,以2为公比的等比数列,
∴an+1-an=3•2n-1,
∴a2-a1=3•20,a3-a2=3•21,a4-a3=3•22,…,an-an-1=3•2n-2,
∴an−a1=
3(1−2n−1)
1−2=3(2n−1−1).即an=3•2n−1−1.(8分)
(3)∵an=3•2n-1-1,
∴sn=3•
1−2n
1−2−n=3•2n-n-3.(12分)
∴
an+2−an+1
an+1−an=2,故数列{an+1−an}是等比数列(4分).
(2)由(1)知数列{an+1-an}以是a2-a1=3为首项,以2为公比的等比数列,
∴an+1-an=3•2n-1,
∴a2-a1=3•20,a3-a2=3•21,a4-a3=3•22,…,an-an-1=3•2n-2,
∴an−a1=
3(1−2n−1)
1−2=3(2n−1−1).即an=3•2n−1−1.(8分)
(3)∵an=3•2n-1-1,
∴sn=3•
1−2n
1−2−n=3•2n-n-3.(12分)
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