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设数列{an}满足an=3an-1+2(n≥2,n∈N+),且a1=2,bn=log3(an+1).(1)证明:数列{an+1}为等比数列;(2)求数列{anbn}的前n项和Sn.
题目详情
设数列{an}满足an=3an-1+2(n≥2,n∈N+),且a1=2,bn=log3(an+1).
(1)证明:数列{an+1}为等比数列;
(2)求数列{anbn}的前n项和Sn.
(1)证明:数列{an+1}为等比数列;
(2)求数列{anbn}的前n项和Sn.
▼优质解答
答案和解析
(1)证明:∵数列{an}满足an=3an-1+2(n≥2,n∈N+),
∴an+1=3(an-1+1),
∴数列{an+1}为等比数列,首项为3,公比为3;
(2) 由(1)可得:an=3n-1.
∵bn=log3(an+1),∴bn=log3(3n+1-1)=n.
∴anbn=n•(3n-1)=n•3n-n.
令Tn=3+2×32+3×33+…+n•3n,
∴3Tn=32+2×33+…+(n-1)•3n+n•3n+1,
∴-2Tn=3+32+…+3n-n•3n+1=
-n•3n+1=
•3n+1-
,
∴Tn=
•3n+1+
.
∴数列{anbn}的前n项和Sn=
•3n+1+
-
.
∴an+1=3(an-1+1),
∴数列{an+1}为等比数列,首项为3,公比为3;
(2) 由(1)可得:an=3n-1.
∵bn=log3(an+1),∴bn=log3(3n+1-1)=n.
∴anbn=n•(3n-1)=n•3n-n.
令Tn=3+2×32+3×33+…+n•3n,
∴3Tn=32+2×33+…+(n-1)•3n+n•3n+1,
∴-2Tn=3+32+…+3n-n•3n+1=
3(3n-1) |
3-1 |
1-2n |
2 |
3 |
2 |
∴Tn=
2n-1 |
4 |
3 |
4 |
∴数列{anbn}的前n项和Sn=
2n-1 |
4 |
3 |
4 |
n(n+1) |
2 |
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