等差数列{An},{Bn}的前n项和分别是Sn和Tn,若Sn/Tn=2n/(3n+1),则lim x→∞(An/Bn)等于___.
人气:256 ℃ 时间:2020-04-22 14:32:25
解答
法一:
An/Bn=[Sn-S(n-1)]/[Tn-T(n-1)]
=[2n-2(n-1)]/{[(3n+1)-[3(n-1)+1]}
=2/3
法二:
因为题目已给出是等差数列,故设
Sn=n*2n,Tn=n*(3n+1)
则An=4n-2,Bn=6n-2
则lim x→∞(An/Bn)=lim x→∞(4n-2)/(6n-2)=4/6=2/3
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