Question

記Sn是等差數(shù)列{an}前n項(xiàng)的和,Tn是等比數(shù)列{bn}前n項(xiàng)的積,設(shè)等差數(shù)列{an}公差d≠0,若對(duì)小于2011的正整數(shù)n,都有Sn=S2011-n成立,則推導(dǎo)出a1006=0.設(shè)等比數(shù)列{bn}的公比q≠1,若對(duì)于小于23的正整數(shù)n,都有Tn=T23-n成立,則(　　)

(A)b11=1 (B)b12=1 (C)b13=1 (D)b14=1

 

Answer 1

B

【解析】由等差數(shù)列中Sn=S2011-n,可導(dǎo)出中間項(xiàng)a1006=0,類比得等比數(shù)列中Tn=T23-n,可導(dǎo)出中間項(xiàng)b12=1.