Question

已知两个等差数列{an},{bn}的前n项和分别是Sn,Tn,若 Sn/Tn =(2n)/(3n+1),则 an/bn=

Answer 1

等差数列数列的性质a1+a[2n-1]=2an

因为S[2n-1]=[(2n-1)(a1+a[2n-1])]/2=(2n-1)an

T[2n-1]=[(2n-1)(b1+b[2n-1])]/2=(2n-1)bn

所以an/bn=S[2n-1]/T[2n-1]=2(2n-1)/[3(2n-1)+1]

=(2n-1)/(3n-1)