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~
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~
答
(2n-1)/(3n-1)
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=1
2b1-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),