解(1)∵f(x)≤0仅有唯一的x值满足,∴△=0,∴a=0或4,∵a≠0,∴a=4
S
n=n
2-4n,a
n=
![](http://thumb.1010pic.com/pic5/latex/51145.png)
=2n-5
(2)b
n:b
1=2×1-5,b
2=2×2-5,b
3=2×4-5,…b
n=2×2
n-1-5
T
n=b
1+b
2+b
3+…+b
n=2(1+2+4+…+2
n-1)-5n
=2
![](http://thumb.1010pic.com/pic5/latex/51146.png)
-5n-2
(3)(理科)
![](http://thumb.1010pic.com/pic5/latex/51147.png)
,∴c
n+2-c
n=2,c
1=1,c
2=4
c
n:1,3,5,7,9…
4,6,8,10…
当n为偶数,n=2k,H
n=5+9+13+…=5k+
![](http://thumb.1010pic.com/pic5/latex/51148.png)
当n为奇数,n=2k-1,H
n=1+(7+11+15+…)
=1+7(k-1)+
![](http://thumb.1010pic.com/pic5/latex/51149.png)
∴H
n=
![](http://thumb.1010pic.com/pic5/latex/51150.png)
当n=2k与n=2k-1时,分别比较H
n与S
n大小(作差比较)
当1≤n≤10时,H
n>S
n当n≥11时,H
n<S
n (4)(文科)c
n=
![](http://thumb.1010pic.com/pic5/latex/51151.png)
c
1=
![](http://thumb.1010pic.com/pic5/latex/8.png)
,c
2=-2,当n≥3时,4n+
![](http://thumb.1010pic.com/pic5/latex/51152.png)
单调递增,且4n+
![](http://thumb.1010pic.com/pic5/latex/51152.png)
-16>0,
∴(c
n)
min=c
2=-2;∴(c
n)
max=c
3=1
分析:(1)根据二次函数性质,,△=0,解方程得出a的值,得出S
n=的解析式,利用数列中Sn与a
n的固有关系an=
![](http://thumb.1010pic.com/pic5/latex/6649.png)
,求出{a
n}的通项
(2)由已知,得出b
n=2×2
n-1-5,采用等比数列求和公式,分组法求和.
(3)(理)由
![](http://thumb.1010pic.com/pic5/latex/51147.png)
得出c
n+2-c
n=2,偶数项成等差数列,奇数项也成等差数列,对n分奇偶性分类求和.
(4)(文)c
n=
![](http://thumb.1010pic.com/pic5/latex/51151.png)
利用函数的数列性质,得出{c
n }的单调性,再求出最值即可.
点评:本题考查构造法求数列通项公式,等比数列的判定,数列公式法、分组法求和,数列的函数性质.考查推理论证、计算能力,分类讨论的思想.