Question

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

Answer 1

∵{an}与{bn}是等差数列

∴Sn=[n(a1+an)]/2

Tn=[n(b1+bn)]/2

∴Sn/Tn=(a1+an)/(b1+bn)

∵等差数列{an}与{bn}的前n项和的比为2n:(3n+1)

∴(a1+an)/(b1+bn)=2n:(3n+1)

假设(n+1)/2 =k {(n+1)/2为项数}

则n=2k-1

则ak/bk = 2(2k-1)/[3(2k-1)+1]

=(2k-1)/(3k-1)

即an/bn =(2n-1)/(3n-1)

或者说：

Sn/Tn=2n/(3n+1),即

S(2n-1)/T(2n-1)=2(2n-1)/[3(2n-1)+1]=(2n-1)/(3n-1),即

[A1+A（2n-1）]/[B1+B(2n-1)]=(2n-1)/(3n-1),即

2An/2Bn=(2n-1)/(3n-1),

An/Bn=(2n-1)/(3n-1)