问题
解答题
已知等差数列{an}的前n项和为An,a1+a5=6,A9=63. (1)求数列{an}的通项公式an及前n项和An; (2)数列{bn}的前n项和Bn满足:6Bn=8bn-1,(n∈N*),数列{an•bn}的前n项和为Sn,求证:
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答案
(1)∵A9=63,∴9a5=63,∴a5=7
∵a1+a5=6,∴a1=-1,∴d=
a5-a1 |
4 |
∴an=2n-3,An=
n(-1+2n-3) |
2 |
(2)证明:∵6Bn=8bn-1,6Bn-1=8bn-1-1,(n≥2,n∈N*)
∴两式相减可得:6bn=8bn-8bn-1
∴
bn |
bn-1 |
∵6b2=8b1-1
∴b1=
1 |
2 |
∴bn=22n-3
∴anbn=(2n-3)•22n-3
∴Sn=-1•2-1+1•21+…+(2n-3)•22n-3
∴4Sn=-1•21+1•23+…+(2n-3)•22n-1
两式相减可得-3Sn=
(11-6n)22n-11 |
6 |
∴
Sn |
4n |
6n-11 |
18 |
11 |
18•4n |
∴
Sn+1 |
4n+1 |
Sn |
4n |
11 |
6×4n+1 |
∴
Sn |
4n |
∴
Sn |
4n |
S1 |
4 |
1 |
8 |