已知数列{an}的前项和为Sn,且满足Sn=
(1)求数列{an}的通项公式; (2)设Tn为数列{
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(本小题满分14分)
(1)当n=1时,a1=S1=2…(2分)
当n≥2时,an=Sn-Sn-1=(
n2+1 2
n)-[3 2
(n-1)2+1 2
(n-1)]=n+1,…(6分)3 2
∵a1=2,∴an=n+1(n∈N*).…(7分)
(2)
=1 anan+1
=1 (n+1)(n+2)
-1 n+1
,…(9分)1 n+2
∴Tn=
-1 2
+1 3
-1 3
+••+1 4
-1 n+1
=1 n+2
-1 2
=1 n+2
…(11分)n 2(n+2)
又Tn>
,得1005 2012
>n 2(n+2)
∴n>2010…(13分)1005 2012
∴n的最小值为2011…(14分)