根據(jù)等差數(shù)列的前n項(xiàng)和是關(guān)于n的沒(méi)有常數(shù)項(xiàng)的一元二次函數(shù)
設(shè)這兩個(gè)等差數(shù)列an和bn的前n項(xiàng)和分別是Sn和Tn
由Sn/Tn=2n/(3n+1)
設(shè)Sn=2kn² Tn=kn(3n+1)
所以當(dāng)n≥2時(shí),
an/bn=[Sn-S(n-1)]/[Tn-T(n-1)]
=[2kn²-2k(n-1)²]/[kn(3n+1)-k(n-1)(3n-2)]
=2[n²-n²+2n-1]/[3n²+n-(3n²-2n-3n+2)]
=2(2n-1)/(6n-2)
=(2n-1)/(3n-1)
當(dāng)n=1時(shí),a1/b1=2/4=1/2,也滿足an/bn=(2n-1)/(3n-1)
綜上：兩數(shù)列第n項(xiàng)之比an/bn=(2n-1)/(3n-1)