设数列{an}的首项a1=1,前n项和Sn满足关系:3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4,5,)
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设数列{an}的首项a1=1,前n项和Sn满足关系:3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4,5,).
求证:数列{an}是等比数列;
求证:数列{an}是等比数列;
∵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.
∴(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.
设数列{an}的首项a1=1,前n项和Sn满足关系:3tSn-(2t+3)Sn-1=3t(t>0,n=2,3,4,5,)
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