本题主要考查了数列的递推式的应用,数列的通项公式和数列的求和问题.应熟练掌握一些常用的数列的求和方法如公式法,错位相减法,叠加法等.
(1)把S
n和S
n+1相减整理求得a
n+1=2a
n+3,整理出3+a
n+1=2(3+a
n),判断出数列{3+a
n}是首相为6,公比为2的等比数列,求得3+a
n,则a
n的表达式可得.
(2)把(I)中的a
n代入b
n,求得其通项公式,进而利用错位相减法求得数列的前n项的和.
(3)设存在满足题意,那么等式两边的奇数和偶数来分析不存在。
解析:(Ⅰ)因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232310750680.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311109863.png)
,
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311140646.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311156612.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311171739.png)
,
所以数列
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232310797560.png)
是等比数列,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311218726.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311249764.png)
,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311000585.png)
.
(Ⅱ)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311312837.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232323113271316.png)
,
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311499962.png)
,①
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232323115141450.png)
,②
①-②得,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232323115461417.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311561834.png)
,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232323110781085.png)
.
(Ⅲ)设存在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311608654.png)
,且
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311639483.png)
,使得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311655519.png)
成等差数列,
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311670614.png)
,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311748912.png)
,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311764588.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311780616.png)
,因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311811458.png)
为偶数,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311951450.png)
为奇数,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823232311764588.png)
不成立,故不存在满足条件的三项.