∵等差数列{an},{bn}的前n项和为Sn与Tn,Sn/Tn=(2n+1)/n∴an/bn=1/2*[a1+a(2n-1)]/1/2*[a1+a(2n-1)]=(2n-1)/2*[a1+a(2n-1)]/(2n-1)/2*[a1+a(2n-1)]=S(2n-1)/T(2n-1)=[2*(2n-1)+1]/(2n-1)=[4n-2+1]/(2n-1)=(4n-1)/(2n-...