已知等比数列{an}的前n项和An=(
(1)求数列{an}的通项公式; (2)求数列{bn}的通项公式; (3)若数列{
|
(1)a1=A1=
-c, a2=A2-A1=(1 3
-c)-(1 9
-c)=-1 3
,a3=A3-A2=(2 9
-c)-(1 27
-c)=-1 9
,2 27
又数列{an}成等比数列,
a1=
=a22 a3
=-4 81 - 2 27
=2 3
-c,1 3
所以 c=1;
又公比q=
=a2 a1
,1 3
所以an=-
×(2 3
) n-1=-2×(1 3
)n,n∈N*.1 3
(2)∵
-Sn
=1(n≥2), S1=b1=1,Sn-1
∴数列{
}是首项为1公差为1的等差数列.Sn
∴
=1+(n-1)×1.Sn
∴Sn=n2.
当n≥2,bn=Sn-Sn-1=n2-(n-1)2=2n-1.
∴bn=2n-1(n∈N*);
(3)Tn=
+1 b1b2
+1 b2b3
+…+1 b3b4 1 bnbn+1
=
+1 1×3
+1 3×5
+…+1 5×7 1 (2n-1)(2n+1)
=
(1-1 2
)+1 3
(1 2
-1 3
)+…+1 5
×1 2 1 (2n-1)(2n+1)
=
(1-1 2
)1 2n+1
=
.n 2n+1
由Tn=
>n 2n+1
得n>1001 2010
,1001 8
故满足Tn>
的最小正整数为126.1001 2010