问题
解答题
数列{an}各项均为正数,其前n项和为Sn,且满足2anSn-
(Ⅰ)求证数列{
(Ⅱ)设bn=
|
答案
(Ⅰ)∵2anSn-
=1a 2n
当n≥2时,2(Sn-Sn-1)Sn-(Sn-Sn-1)2=1,
整理得,
-S 2n
=1(n≥2),(2分)S 2n-1
又
=1,(3分)S 21
∴数列{
}为首项和公差都是1的等差数列.(4分)S 2n
∴
=n,又Sn>0,∴Sn=S 2n
(5分)n
∴n≥2时,an=Sn-Sn-1=
-n
,n-1
又a1=S1=1适合此式 (6分)
∴数列{an}的通项公式为an=
-n
(7分)n-1
(Ⅱ)∵bn=
=2 4
-1S 4n
=2 (2n-1)(2n+1)
-1 2n-1
(8分)1 2n+1
∴Tn=
+1 1×3
+…+1 3×5 1 (2n-1)(2n+1)
=1-
+1 3
-1 3
+…+1 5
-1 2n-1 1 2n+1
=1-
=1 2n+1
(10分)2n 2n+1
∴Tn≥
,依题意有2 3
>2 3
(m2-3m),解得-1<m<4,1 6
故所求最大正整数m的值为3 (12分)