(1)当n≥2时,an=Sn-Sn-1=nan-(n-1)an-1-2(n-1)…(2分)
an-an-1=2(n≥2),
数列{an}是以a1=1为首项,以2为公差的等差数列
∴an=2n-1…(6分)
(2)数列{
}的前n项和为Tn,1
anan+1
…(10分)
Tn=
+1
a1a2
+…+1
a2a3
=1
anan+1
+1 1×3
+…+1 3×5
1 (2n?1)×(2n+1) =
[(1 2
?1 1
)+(1 3
?1 3
)+(1 5
?1 5
)+…+(1 7
?1 2n?1
)]1 2n+1 =
(1?1 2
)=1 2n+1
n 2n+1
∴
>n 2n+1
,即n>100 209
,100 9
∴满足Tn>
的最小正整数n是12…(12分)100 209