Question

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

Answer 1

(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)