若等差数列{an}和{bn}的前几项和为Sn和Tn,若Sn/Tn=2n-1/3-n,求an/bn的极限

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若等差数列{an}和{bn}的前几项和为Sn和Tn,若Sn/Tn=2n-1/3-n,求an/bn的极限
若等差数列{an}和{bn}的前几项和为Sn和Tn,若Sn/Tn=2n-1/3-n,求an/bn的极限

若等差数列{an}和{bn}的前几项和为Sn和Tn,若Sn/Tn=2n-1/3-n,求an/bn的极限
Sn=a1n+(n-1)n*d1/2
Tn=b1n+(n-1)n*d2/2
sn/Tn=(a1+an)n/2/(b1+bn)n/2
=(a1+an)/(b1+bn)
=(2n-1)/(3-n)
所以
liman/bn=lim(a1+an)/(b1+bn)
=lim(2n-1)/(3-n)
=-2

由等差数列{an}、{bn}的前n项和分别为Sn和Tn,Sn/Tn=2n/(3n+1)可Tn=(3n+1)*(hn+m)=(3n+1)*(hn-h)=h(3n+1)*(n-1) 则an=Sn

2

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