题目内容
(本题满分12分)已知数列{an}的前n项和为Sn,且an=
(3n+Sn)对一切正整数n成立
(1)证明:数列{3+an}是等比数列,并求出数列{an}的通项公式;
(2)设
,求数列
的前n项和Bn;
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113631333.png)
(1)证明:数列{3+an}是等比数列,并求出数列{an}的通项公式;
(2)设
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113646649.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113677490.png)
解:(1)由已知得Sn=2an-3n,
Sn+1=2an+1-3(n+1),两式相减并整理得:an+1=2an+3
所以3+ an+1=2(3+an),又a1=S1=2a1-3,a1=3可知3+ a1=6
,进而可知an+3![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113693324.png)
所以
,故数列{3+an}是首相为6,公比为2的等比数列,
所以3+an=6
,即an=3(
)
(2)![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113849797.png)
设
(1)
(2)
由(2)-(1)得![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139581027.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139741081.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941140051791.png)
Sn+1=2an+1-3(n+1),两式相减并整理得:an+1=2an+3
所以3+ an+1=2(3+an),又a1=S1=2a1-3,a1=3可知3+ a1=6
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113693324.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113693324.png)
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113787716.png)
所以3+an=6
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113802398.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113818408.png)
(2)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823194113849797.png)
设
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139271029.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139431192.png)
由(2)-(1)得
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139581027.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941139741081.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231941140051791.png)
略
![](http://thumb.zyjl.cn/images/loading.gif)
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