{an}和{bn}是等差数列,(a1+a2+a3+.an)/(b1+b2+b3+.bn)=(3n+1)/(4n+3),对

设Sn=a1+a2+a3+.an
Tn=b1+b2+b3+.bn
则Sn/Tn=(3n+1)/(4n+3),
根据等差数列的前n项和的性质：
设 Sn=kn(3n+1)
Tn=kn(4n+3)
当n≥2时,
an/bn
=[Sn-S(n-1)]/[Tn-T(n-1)]
=[kn(3n+1)-k(n-1)(3n-2)]/[kn(4n+3)-k(n-1)(4n-1)]
=(6n-2)/(8n-1)
当n=1时,a1/b1=(3×1+1)/(4×1+3)=4/7,也满足上式
故an/bn=(6n-2)/(8n-1)