(1)设 =x得:(1-b)x2+cx+a=0,由根与系数的关系,得:,
解得 ,代入解析式 f(x)=,由 f(-2)=<-,
得c<3,又c∈N,b∈N,若c=0,b=1,则f(x)=x不止有两个不动点,∴c=2,b=2,于是f(x)=,(x≠1).
(2)由题设,知 4Sn•=1,所以,2Sn=an-an2①;
且an≠1,以n-1代n得:2Sn-1=an-1-an-12,②;
由①-②得:2an=(an-an-1)-(an2-an-12),即(an+an-1)(an-an-1+1)=0,
∴an=-an-1或an-an-1=-1,以n=1代入①得:2a1=a1-a12,
解得a1=0(舍去)或a1=-1;由a1=-1,若an=-an-1得a2=1,这与an≠1矛盾,
∴an-an-1=-1,即{an}是以-1为首项,-1为公差的等差数列,∴an=-n;
(3)由an=-n,知(1-)an+1=(1+)-(n+1)=()n+1,
(1-)an=(1+)-n=()n,
当n=1时,()n+1=,()n=,()n+1<<()n成立.
假设n=k时,()k+1<<()k成立,
则当n=k+1时,()k+2<<()k+1成立.
所以,(1-)an+1<<(1-)an.