Question

一道等差数列的题目不会做题目：设等差数列{an},{bn}的前n项和分别为Sn和Tn.（1）求证：an/bn=(S2n-

Answer 1

(1)证明：令{an}的公差为d1,{bn}的公差为d2.

则S2n-1=[a1+a1+(2n-1-1)d1](2n-1)/2=2[a1+(n-1)d1](2n-1)/2

T2n-1=[b1+b1+(2n-1-1)d2](2n-1)/2=2[b1+(n-1)d2](2n-1)/2

(S2n-1)/(T2n-1)=[a1+(n-1)d1]/[b1+(n-1)d2]=an/bn

（2）由上知a3/b3=S5/T5=（5n+1)/(3n-1)=13/7.