Question

设数列(an)de 首项a1=1,前n项和Sn满足3tSn-(2t+3).Sn-1=3t(t>0,n=1,2,3,4,...)证明数列an是等比数列

Answer 1

由3tSn-(2t+3).S(n-1)=3t得
3tS(n-1)-(2t+3).S(n-2)=3t
两式相减得
3t*（Sn-S(n-1)）-(2t+3)*(S(n-1)-S(n-2))=0
即得
3t*an-(2t+3)*a(n-1)=0
所以
an/a(n-1)=(2t+3)/(3t)=(常数)
最后，数列{an}是等比数列,首项为1，共比为(2t+3)/(3t)

Answer 2

∵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。