設(shè)數(shù)列{an}的前n項和為Sn,a1=1且對于任意正整數(shù)n,點(an+1,Sn)在直線2x+y-2=0上.
(1)求數(shù)列{an}的通項公式;
(2)設(shè)數(shù)列{bn}滿足:bn=nan,Tn為數(shù)列{bn}的前n項和,求證:當n≥2時,Tn<4.
【答案】
分析:(1)根據(jù)點在直線上則S
n=2-2a
n+1,根據(jù)遞推關(guān)系可得n≥2時,S
n-1=2-2a
n,兩式作差,可得數(shù)列{a
n}是以1為首項,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/0.png)
為公比的等比數(shù)列,從而通項公式;
(2)根據(jù)數(shù)列{b
n}的特征,利用錯位相消法求出其前n項和T
n,然后化簡整理可證得結(jié)論,當n≥2時,T
n<4.
解答:(本小題滿分12分)
解:(1)點(a
n+1,S
n)在直線2x+y-2=0上
∴2a
n+1+S
n-2=0即∴S
n=2-2a
n+1 ①
當n≥2時,∴S
n-1=2-2a
n ②…(3分)
由①-②可得:a
n=2a
n+1∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/1.png)
(n≥2)又a
1=1,a
2=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/2.png)
符合上式
數(shù)列{a
n}是以1為首項,
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/3.png)
為公比的等比數(shù)列
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/4.png)
…(6分)
(2)由(1)知b
n=na
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/5.png)
∴T
n=1+2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/6.png)
+3
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/7.png)
+4
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/8.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/9.png)
…③
∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/10.png)
T
n=
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/11.png)
+2
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/12.png)
+3
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/13.png)
+4
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/14.png)
+…+
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/15.png)
…④
由③-④得∴
![](http://thumb.zyjl.cn/pic6/res/gzsx/web/STSource/20131025125223218116120/SYS201310251252232181161018_DA/16.png)
…(12分)
點評:本題主要考查了等比數(shù)列的通項公式,數(shù)列的求和,以及不等式的證明,是一道綜合題,屬于中檔題.