(1)由S
n+1=aS
n+2(n=1,2,…,2k-1) (1)
S
n=aS
n-1+2(n=2,3,…,k) (2)……………………………2分
(1)-(2)得a
n+1=a·a
n(n=2,3,…,2k-1)
由(1)式S
2=aS
1+2,a
1+a
2=aS
1+2……………………………………………………3分
解得a
2=2a,因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235016304.gif)
所以{a
n}是以2为首项,a为公比的等比数列,a
n=2·a
n-1(n=1,2…,2k)…………4分
(2)∵b
n-b
n-1=log
2a
n-log
2a
n-1=log
2a
n-1log
2![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235032380.gif)
=log
2a (n=2,3…,2k)
∴{b
n}是以b
1=1为首项,以log
2a(a>1)为公差的等差数列………………………6分
∴T
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235063449.gif)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235094498.gif)
=n+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140234938440.gif)
(a>1,n=1,2,…,2k)……………8分
(3)c
n=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235110216.gif)
=1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235125468.gif)
=1+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235250341.gif)
(n=1,2,…,2k)……………………………10分
当c
n≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235266192.gif)
时, n≤k+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235281193.gif)
,n为正整数,知n≤k时,c
n<
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235297219.gif)
当n≥k+1时,c
n>
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235297219.gif)
……………………………………………………………………11分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235344790.gif)
=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235359192.gif)
-c
1)+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235359192.gif)
-c
2)+…+(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235359192.gif)
-c
k)+(c
k+1-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235359192.gif)
)+…+(c
2
k-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235359192.gif)
)
=(c
k+1+c
k+2+…+c
2k)-(c
1+c
2+…+c
k)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235437265.gif)
{[k+(k+1)+…+(2k-1)]+2k}-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235453220.gif)
{[1+2+…+(k-1)]+k}
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235437265.gif)
[
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235484387.gif)
-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235500421.gif)
]
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235515405.gif)
≥
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140235515273.gif)
即11k
2-72k+3
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/2014082314023553172.gif)
6≥0,(11k-6)(k-6)≥0解得k≥6或k≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823140234907219.gif)
所以满足条件的k的最小值为6…………………………14分