数列{an}的前n项为Sn,Sn=2an-3n(n∈N*). (1)证明:数列{an+3}是等比数列; (2)求数列{an}的通项公式an.

问题描述:

数列{an}的前n项为Sn,Sn=2an-3n(n∈N*).
(1)证明:数列{an+3}是等比数列;
(2)求数列{an}的通项公式an

(1)证明:由Sn=2an-3n,得Sn-1=2an-1-3(n-1)(n≥2),则有an=2an-2an-1-3an+3=2(an-1+3)(n≥2),∵a1=S1=2a1-3,∴a1=3,∴a1+3=6≠0,由此可得a2+3=12≠0,以此类推an+3≠0,∴an+3an−1+3=2(n≥2),∴数...