(I)因为曲线在pn处的切线与AAn平行 ∴6xn=⇒2xn=an+2 (Ⅱ)∵xn+1=[f(xn-1)+1]+1 ∴xn+1=[3(xn-1)2-1+1]+1,⇒xn+1=t(xn-1)2+1 从而logt(xn+1-1)=1+2logt(xn-1)⇒logt(xn+1-1)+1=2[logt(xn-1)+1] ∴{logt(xn-1)+1}是一个公比为2的等比数列 (III)由(II)知:logt(xn-1)+1=(logt2+1)2n-1 ∴xn=1+(2t)2n-1,从而an=2xn-2=(2t)2n-1 ∴an+1<an,∴(2t)2n<(2t)2n-1 ∴0<2t<1⇒0<t< |