等差数列an`bn`的前n项和分别为Sn`Tn.若Sn/Tn=(7n+1)/(4n+27),求an/bn
问题描述:
等差数列an`bn`的前n项和分别为Sn`Tn.若Sn/Tn=(7n+1)/(4n+27),求an/bn
答
(2n-1)an=S(2n-1),
(2n-1)bn=T(2n-1),
所以an/bn=S(2n-1)/T(2n-1)
=[7(2n-1)+1]/[4(2n-1)+27]
=(14n-6)/(8n+23)