(1)
2tSn-(2t+1)S(n-1)=2t (1)
2tS(n-1)-(2t+1)S(n-2)=2t (2)
(1)-(2)
2tan-(2t+1)a(n-1) = 0
an/a(n-1) = (2t+1)/2t
=>{an}为等比数列
(2)
数列{an}首项a1=1,前n项和Sn满足等式2tSn-(2t+1)S(n-1)=2t(常数t>0,n=2,3,4···
2个回答
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