问题
解答题
已知函数f(x)=x2+(a-1)x+b+1,当x∈[b,a]时,函数f(x)的图象关于y轴对称,数列{an}的前n项和为Sn,且Sn=f(n). (1)求数列{an}的通项公式; (2)设bn=
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答案
(1)∵函数f(x)的图象关于y轴对称,
∴a-1=0,且a+b=0,解得a=1,b=-1,
∴Sn=n2,即有an=Sn-Sn-1=2n-1(n≥2).
a1=S1=1也满足,
∴an=2n-1.(5分)
(2)由(1)得bn=
,2n-1 2n
∴Tn=
+1 21
+3 22
+…+5 23
+2n-3 2n-1
,①2n-1 2n
Tn=1 2
+1 22
+…+3 23
+2n-5 2n-1
+2n-3 2n
,②2n-1 2n+1
①-②得
Tn=1 2
+1 2
+2 22
+…+2 23
+2 2n-1
-2 2n 2n-1 2n+1
=
+(1 2
+1 2
+1 22
+…+1 23
)-1 2n-1 2n-1 2n+1
=
-3 2
-1 2n-1
,2n-1 2n+1
∴Tn=3-
-1 2n-2
=3-2n-1 2n
.(9分)2n+3 2n
设g(n)=
,n∈N+,2n+3 2n
则由
=g(n+1) g(n)
=2n+5 2n+1 2n+3 2n
=2n+5 2(2n+3)
+1 2
≤1 2n+3
+1 2
<1,得g(n)=1 5
(n∈N+)随n的增大而减小,2n+3 2n
∴g(n)≤g(1),
即Tn≥3-
=2+3 2
.1 2
又Tn>m恒成立,
∴m<
.(12分)1 2