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两个等差数列{an},{bn}的前n项和分别为Sn,Tn,若Sn/Tn=2n/3n+1,求an/bn.
人气:318 ℃ 时间:2020-05-23 09:26:41
解答
S(2n-1)=(A1+A(2n-1))×(2n-1)/2
=(A1+A1+(2n-2)d)×(2n-1)/2
=(A1+(n-1)d)×(2n-1)
=An×(2n-1)
同理
T(2n-1)=Bn×(2n-1)
[An×(2n-1)]/[Bn×(2n-1)]
=S(2n-1)/T(2n-1)
=2(2n-1)/[3(2n-1)+1]
=(4n-2)/(6n-3+1)
=(2n-1)/(3n-1)
An/Bn=(2n-1)/(3n-1)
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