试题分析:(I){a
n}是一等比数列,且a
1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
.设等比数列{a
n}的公比为q,由S
1+a
1,S
2+a
2,S
3+a
3成等差数列,可得一个含公比q的方程,解这个方程便得公比q,从而得数列{a
n}通项公式. (Ⅱ)由题设及(I)可得:b
n=a
nlog
2a
n=-n?(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
)
n,由等差数列与等比数列的积或商构成的新数列,求和时用错位相消法.
试题解析:(I)设等比数列{a
n}的公比为q,由题知 a
1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
,
又∵ S
1+a
1,S
2+a
2,S
3+a
3成等差数列,
∴ 2(S
2+a
2)=S
1+a
1+S
3+a
3,
变形得S
2-S
1+2a
2=a
1+S
3-S
2+a
3,即得3a
2=a
1+2a
3,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308757389.png)
q=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
+q
2,解得q=1或q=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
, 4分
又由{a
n}为递减数列,于是q=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
,
∴a
n=a
1![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308679407.png)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
)
n. 6分
(Ⅱ)由于b
n=a
nlog
2a
n=-n?(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694339.png)
)
n,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240333088661511.png)
,
于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240333088811364.png)
,
两式相减得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240333088971258.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240333089131207.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824033308694925.png)
. 12分