题目

(理)已知数列{an}的前n项和为Sn,且满足a1=,an+2SnSn-1=0(n≥2),(1)判断{}是否为等差数列?并证明你的结论;(2)求Sn和an;(3)求证:S12+S22+…+Sn2≤.(文)数列{an}的前n项和Sn(n∈N*),点(an,Sn)在直线y=2x-3n上.(1)求证:数列{an+3}是等比数列;(2)求数列{an}的通项公式;(3)数列{an}中是否存在成等差数列的三项?若存在,求出一组适合条件的三项;若不存在,说明理由. 答案：(理)(1)解:S1=a1=,∴=2.                                               当n≥2时,an=Sn-Sn-1,即Sn-Sn-1=-2SnSn-1,                                         ∴=2.故{}是以2为首项,以2为公差的等差数列.                                  (2)解:由(1)得=2+(n-1)·2=2n,Sn=.                                         当n≥2时,an=-2SnSn-1=;                                            当n=1时,a1=.∴an=                                                  (3)证法一:①当n=1时,成立.                                ②假设n=k时,不等式成立,即≤成立.则当n=k+1时,≤==＜=,即当n=k+1时,不等式成立.由①②可知对任意n∈N*不等式成立.                                           证法二:==≤==.(文)(1)证明:由题意知Sn=2an-3n,∴an+1=Sn+1-Sn=2an+1-3(n+1)-2an+3n.∴an+1=2an+3.                                                                ∴an+1+3=2(an+3).∴=2.又a1=S1=2a1-3,a1=3,∴a1+3=6.                                                                  ∴数列{an+3}成以6为首项以2为公比的等比数列.                              (2)解:由(1)得an+3=6·2n-1=3·2n,∴an=3·2n-3.                                                                 (3)解:设存在s、p、r∈N*且s＜p＜r使as、ap、ar成等差数列,∴2ap=as+ar.∴2(3·2p-3)=3·2s-3+3·2r-3.∴2p+1=2s+2r,                                                                即2p-s+1=1+2r-s.                                                               (*)∵s、p、r∈N*且s＜p＜r,∴2p-s+1为偶数,1+2r-s为奇数.∴(*)为矛盾等式,不成立.故这样的三项不存在.

查看本题答案和解析