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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;
人气:442 ℃ 时间:2020-03-18 01:14:22
解答
∵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}...
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