已知等差数列{an}，{bn}的前n项和分别为An、Bn，且满足AnBn=2nn+3，则a1+a2+a12b2+b4+b9=．

Question

已知等差数列{a_n}，{b_n}的前n项和分别为A_n、B_n，且满足 (An)/(Bn)= (2n)/(n+3)，则 (a1+a2+a12)/(b2+b4+b9)=___．

Answer 1

∵等差数列{a_n}，{b_n}的前n项和分别为A_n、B_n，它们的公差分别为d、d′，且满足(An)/(Bn)=(2n)/(n+3)，
则(a1+a2+a12)/(b2+b4+b9)=(3a1+12d)/(3b1+12d′)=(3a5)/(3b5)=(a5)/(b5)=((9)/(2)(a1+a9))/((9)/(2)(b1+b9))=(A9)/(B9)=(2×9)/(9+3)=(3)/(2)，
故答案为：(3)/(2)．