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解法一：直接求an和bn
设{an}首项为a1,公差为d1,{bn}首项为b1,公差为d2,则当n=1时得
S1/T1=2/4=1/2,即b1=2a1
当n=2时,得S2/T2=4/7,即(2a1+d1)/(2b1+d2)=4/7,得7d1=2a1+4d2
当n=3时,得5d1=a1+3d2,因此d1=2a1,d2=3a1,则
`<n→∞>lim(an/bn)
=<n→∞>lim{[a1+(n-1)d1]/[b1+(n-1)d2]}
=<n→∞>lim[(2n-1)/(3n-1)]=2/3

解法二：由等差数列的通项公式an=[a1+a(2n-1)]/2,bn=[b1+b(2n-1)]/2,
所以an/bn=[(a1+a(2n-1))(2n-1)/2]/[(b1+b(2n-1))(2n-1)/2]
`````````=S(2n-1)/S(tn-1)
`````````=2(2n-1)/[3(2n-1)+1]
`````````=(4n-2)/(6n-2)
所以<n→∞>lim(an/bn)=2/3

解法三：等差数列前n项和公式Sn=na1+n(n-1)d/2,当d≠0时是关于n的不含常数项的二次式,所以Sn/Tn=2n·nk/[(3n+1)kn] (k为常数).
令Sn=2kn²,则Tn=3kn²+kn,所以a1=S1=2k,b1=T1=4k,
当n≥2时,an=Sn-S(n-1)=(4n-2)k,bn=Tn-T(n-1)=(6n-2)k.
故<n→∞>lim(an/bn)
`=<n→∞>lim(an/bn)[(4n-2)k/(6n-2)k]
`=2/3.