數(shù)列{an}中,a3=1,a1+a2+…+an=an+1(n∈N*).
(Ⅰ)求a1,a2,a4,a5;
(Ⅱ)求數(shù)列{an}的前n項(xiàng)和Sn;
(Ⅲ)設(shè)bn=log2Sn,存在數(shù)列{cn}使得cn•bn+3•bn+4=n(n+1)(n+2)Sn,試求數(shù)列{cn}的前n項(xiàng)和Tn.
【答案】
分析:(Ⅰ)依題意,可求得a
1=a
2;而a
1+a
2=a
3=1,從而可求a
1,a
2,繼而可求得a
4,a
5;
(Ⅱ)可求得2S
n=S
n+1,即{S
n}是首項(xiàng)為S
1=a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/0.png)
,公比為2的等比數(shù)列,從而可求得S
n=2
n-2;
(Ⅲ)依題意,可求得c
n=n•2
n-2,利用錯(cuò)位相減法即可求得數(shù)列{c
n}的前n項(xiàng)和T
n.
解答:解:(Ⅰ)當(dāng)n=1時(shí),有a
1=a
2;當(dāng)n=2時(shí),有a
1+a
2=a
3;…
∵a
3=1,
∴a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/1.png)
,a
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/2.png)
,a
4=2,a
5=4.…(4分)
(Ⅱ)∵S
n=a
n+1=S
n+1-S
n,…(6分)
∴2S
n=S
n+1∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/3.png)
=2…(8分)
∴{S
n}是首項(xiàng)為S
1=a
1=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/4.png)
,公比為2的等比數(shù)列.
∴S
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/5.png)
•2
n-1=2
n-2…(10分)
(Ⅲ)由S
n=2
n-2,得b
n=n-2,
∴b
n+3=n+1,b
n+4=n+2,
∵c
n•b
n+3•b
n+4=n(n+1)(n+2)S
n,
∴c
n•(n+1)(n+2)=n(n+1)(n+2)2
n-2,
即c
n=n•2
n-2. …(12分)
T
n=1×2
-1+2×2
+3×2
1+4×2
2+…+n•2
n-2…①
則2T
n=1×2
+2×2
1+3×2
2+…+(n-1)•2
n-2+n•2
n-1…②
②一①得
T
n=n•2
n-1-2
-1-2
-2
1-…-2
n-2=n•2
n-1-
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/6.png)
=n•2
n-1+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025124754536135698/SYS201310251247545361356021_DA/7.png)
.…(14分)
點(diǎn)評(píng):本題考查數(shù)列的求和,考查等比數(shù)列的判定,突出考查錯(cuò)位相減法求和,考查等價(jià)轉(zhuǎn)化思想與推理運(yùn)算能力,屬于難題.