已知数列{an}满足:a1=2,an+1=2an+2n+1(n∈N*)(1)记bn=an2n,证明:数列{bn}为等差数列.(2)求数
已知数列{an}满足:a1=2,an+1=2an+2n+1(n∈N*)(1)记bn=an2n,证明:数列{bn}为等差数列.(2)求数列{an}的通项公式an及前n项和为...
已知数列{an}满足:a1=2,an+1=2an+2n+1(n∈N*)(1)记bn=an2n,证明:数列{bn}为等差数列.(2)求数列{an}的通项公式an及前n项和为Sn.
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(1)由已知有:
=
+1(n∈N*),
即:bn+1-bn=1(n∈N*)
∴数列{bn}为等差数列,其首项为1,公差为1
(2)由(1)知:bn=1+(n-1)×1=n(n∈N*)
即:
=n(n∈N*)∴an=n2n(n∈N*)
∴Sn=1×21+2×22+3×23+…+n2n
2Sn=1×22+2×23+…+(n-1)×2n+n2n+1
两式相减,得:
?Sn=21+22+23+…+2n?n2n+1=
?n2n+1=2n+1?2?n2n+1
∴an=n2nSn=(n-1)2n+1+2
an+1 |
2n+1 |
an |
2n |
即:bn+1-bn=1(n∈N*)
∴数列{bn}为等差数列,其首项为1,公差为1
(2)由(1)知:bn=1+(n-1)×1=n(n∈N*)
即:
an |
2n |
∴Sn=1×21+2×22+3×23+…+n2n
2Sn=1×22+2×23+…+(n-1)×2n+n2n+1
两式相减,得:
?Sn=21+22+23+…+2n?n2n+1=
2(1?2n) |
1?2 |
∴an=n2nSn=(n-1)2n+1+2
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