题目内容
(本小题满分12分)已知数列{
}满足
=
,
是{
}的前
项的和,
. (1)求
;(2)证明:![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800659665.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800425212.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800441220.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800456579.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800441220.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800425212.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800503192.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800628247.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800441220.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800659665.gif)
(1)Sn=
(2)略
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800706600.gif)
(1)由题意Sn=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800722716.gif)
两式相减得2an+1=(n+1)an+1-nan即(n-1)an+1=nan,
所以
再相加得
即![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800768483.gif)
所以数列{an}是等差数列. ………………4分
∵a1=
∴a1=0,
又a2=1,则公差为1,∴an=n-1,
所以数列{an}的前n项的和为Sn=
………………6分
(2)(1+![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231418008311776.gif)
………………8分
①当n=1时:(1+![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800846484.gif)
②当n≥2时:
∵
………………10分
∴(1+
…+![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800893935.gif)
而(1+
,∴![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800924657.gif)
综上所证:
………………12分
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800722716.gif)
两式相减得2an+1=(n+1)an+1-nan即(n-1)an+1=nan,
所以
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800737579.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800753574.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800768483.gif)
所以数列{an}是等差数列. ………………4分
∵a1=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800784275.gif)
又a2=1,则公差为1,∴an=n-1,
所以数列{an}的前n项的和为Sn=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800706600.gif)
(2)(1+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231418008311776.gif)
………………8分
①当n=1时:(1+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800846484.gif)
②当n≥2时:
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408231418008621530.gif)
∴(1+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800878622.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800893935.gif)
而(1+
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800909747.gif)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800924657.gif)
综上所证:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823141800940563.gif)
![](http://thumb.zyjl.cn/images/loading.gif)
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