问题描述: 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 1个回答 分类:数学 2014-12-07 问题解答: 我来补答 ∵{an}与{bn}是等差数列∴Sn=[n(a1+an)]/2Tn=[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) 展开全文阅读