问题描述: 等差数列{an} {bn}的前n项的分别为Sn Tn.若Sn/Tn=2n/(3n+1),求an/bn的表达式.嘛烦各位大侠了~ 1个回答 分类:数学 2014-12-12 问题解答: 我来补答 Sn/Tn=2n/(3n+1)(a1+a1+(n-1)*d1)/(b1+b1+(n-1)*d2)=2n/(3n+1)(2a1-d+n*d1)/(2b1-d2+n*d2)=2n/(3n+1)->2a1=d1,d1=2,->a1=12b1-d2=-1,d2=3->b1=1,an=(1+(n-1)*2)=2n-1,bn=(1+(n-1)*3)=3n-2,an/bn=(2n-1)/(3n-2), 展开全文阅读