(1)∵Sn=A+Ban+C,A=0,B=3,C=-2, ∴Sn=3an-2, ∴当n=1时,a1=3a1-2,解得a1=1; 当n≥2时,an=Sn-Sn-1=3an-3an-1, 整理,得2an=3an-1, ∴=, ∴an=()n-1. (2)∵Sn=A+Ban+C,A=1,B=,C=, ∴Sn=+an+, ∴当n=1时,a1=+a1+,解得a1=, 当n≥2时,an=Sn-Sn-1=-+an-an-1 整理,得(an+an-1)(an-an-1-)=0, ∵an>0,∴an-an-1=, ∴{an}是首项为,公差为的等差数列, ∴Sn=+=. (3)若数列{an}是公比为q的等比数列, ①当q=1时,an=a1,Sn=na1 由Sn=A+Ban+C,得na1=A+Ba1+C恒成立 ∴a1=0,与数列{an}是等比数列矛盾; ②当q≠±1,q≠0时,an=a1qn-1,Sn=qn-, 由Sn=A+Ban+C恒成立, 得A××q2n+(B×-)×qn+C+=0对于一切正整数n都成立 ∴A=0,B=≠1或或0,C≠0, 事实上,当A=0,B≠1或或0,C≠0时, Sn=Ban+Ca1=≠0, n≥2时,an=Sn-Sn-1=Ban-Ban-1, 得=≠0或-1 ∴数列{an}是以为首项,以为公比的等比数列. |