(1)当n=1时,a 1 =S 1 =3×1-1=2; 当n≥2时,a n =S n -S n-1 = (
∴ a n =
∵n=1时,a 1 =S 1 =3×1-1=2不满足 a n =(
∴{a n }不是等比数列; (2)∵ b n =
∴
∴数列 {
∴
两式相减可得
∴T n = 6-2(n+3)(
(3)由(2)有b n+1 -b n =
∴n≤2时,有b n+1 -b n ≤0;n>2时,b n+1 -b n >0 ∴b n 的最小值为b 2 =b 3 =
∴-
∴t 2 -2t-3>0 ∴t>3或t<-1 ∴t的最小正整数值是4. |