∵3t*Sn-(2t+3)S(n-1)=3t,3t*[S(n-1)+an]-(2t+3)S(n-1)=3t, ∴(t-3)S(n-1)+3tan=3t…①,(t-3)Sn+3ta(n+1)=3t…②, ②-①得,(t-3)[Sn-S(n-1)]+3t[a(n+1)-an]=0=(t-3)(an)+3t[a(n+1)-an]=0, ∴a(n+1)/an=(2t+3)/3t,∴{an}是等比数列。
