an等差数列 bn前n项和sn满足sn=3(bn-1)/2 且a2=b1 a5=b2 ⑴求an bn通项 ⑵设tn为数列sn的前n项和,求tn
人气:304 ℃ 时间:2019-08-19 03:09:55
解答
1.Sn=3/2bn-3/2S(n-1)=3/2b(n-1)-3/2bn=Sn-S(n-1)=3/2bn-3/2b(n-1)bn=3b(n-1)所以{bn}是等比数列,公比3而b1=S1=3/2b1-3/2b1=3所以bn=3^na2=b1=3,a5=b2=9d=(a5-a2)/3=2an=2n-12.Sn=3/2*3^n-3/2Tn=9/2(1-3^n)/(1-3)-3/...
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