Question

已知两个等差数列{An} {bn},他们的前n项和分别是Sn,Tn ,若Sn/Tn=2n+3/3n-1,求an/bn

Answer 1

因为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)/T(2n-1)

=[2(2n-1)+3]/[3(2n-1)-1]

=(4n+1)/(6n-4).