已知各项均不为零的数列{an}满足an+12=anan+2，且32a8-a3=0，记Sn是数列{an}的前n项和，则S6a1-S3的值为（）A.-218B.218C.-9D.9

Question

已知各项均不为零的数列{a_n}满足a_n+1²=a_na_n+2，且32a₈-a₃=0，记S_n是数列{a_n}的前n项和，则

(S6)/(a1-S3)的值为（　　）

A. -

(21)/(8)

B.

(21)/(8)

C. -9

D. 9

Answer 1

各项均不为零的数列{a_n}满足a_n+1²=a_na_n+2，∴此数列是等比数列．设公比为q．
∵32a₈-a₃=0，∴32a3q5-a3=0，解得q=(1)/(2)．
则(S6)/(a1-S3)=((a1(1-q6))/(1-q))/(a1-(a1(1-q3))/(1-q))=((1+q3)(1-q3))/(-q(1-q)(1+q))=-((1+(1)/(23))(1-(1)/(23)))/(-(1)/(2)×(1-(1)/(2))×(1+(1)/(2)))=-(21)/(8)．
故选：A．