Question

(理)已知數(shù)列{a_n}的前n項(xiàng)和為S_n,且滿(mǎn)足a₁=,a_n+2S_nS_n-1=0(n≥2),

(1)判斷{}是否為等差數(shù)列?并證明你的結(jié)論;

(2)求S_n和a_n;

(3)求證:S₁²+S₂²+…+S_n²≤.

(文)數(shù)列{a_n}的前n項(xiàng)和S_n(n∈N^*),點(diǎn)(a_n,S_n)在直線y=2x-3n上.

(1)求證:數(shù)列{a_n+3}是等比數(shù)列;

(2)求數(shù)列{a_n}的通項(xiàng)公式;

(3)數(shù)列{a_n}中是否存在成等差數(shù)列的三項(xiàng)?若存在,求出一組適合條件的三項(xiàng);若不存在,說(shuō)明理由.

Answer 1

(理)(1)解:S₁=a₁=,∴=2.                                              

當(dāng)n≥2時(shí),a_n=S_n-S_n-1,即S_n-S_n-1=-2S_nS_n-1,                                        

∴=2.

故{}是以2為首項(xiàng),以2為公差的等差數(shù)列.                                  

(2)解:由(1)得=2+(n-1)·2=2n,S_n=.                                        

當(dāng)n≥2時(shí),a_n=-2S_nS_n-1=;                                           

當(dāng)n=1時(shí),a₁=.

∴a_n=                                                 

(3)證法一:①當(dāng)n=1時(shí),成立.                               

②假設(shè)n=k時(shí),不等式成立,即≤成立.

則當(dāng)n=k+1時(shí),

≤

=

=

＜

=,

即當(dāng)n=k+1時(shí),不等式成立.

由①②可知對(duì)任意n∈N^*不等式成立.                                          

證法二:

=

=

≤

=

=.

(文)(1)證明:由題意知S_n=2a_n-3n,

∴a_n+1=S_n+1-S_n=2a_n+1-3(n+1)-2a_n+3n.

∴a_n+1=2a_n+3.                                                               

∴a_n+1+3=2(a_n+3).

∴=2.

又a₁=S₁=2a₁-3,a₁=3,

∴a₁+3=6.                                                                 

∴數(shù)列{a_n+3}成以6為首項(xiàng)以2為公比的等比數(shù)列.                             

(2)解:由(1)得a_n+3=6·2^n-1=3·2ⁿ,

∴a_n=3·2ⁿ-3.                                                                 

(3)解:設(shè)存在s、p、r∈N^*且s＜p＜r使a_s、a_p、a_r成等差數(shù)列,

∴2a_p=a_s+a_r.

∴2(3·2^p-3)=3·2^s-3+3·2^r-3.

∴2^p+1=2^s+2^r,                                                               

即2^p-s+1=1+2^r-s.                                                               (*)

∵s、p、r∈N^*且s＜p＜r,

∴2^p-s+1為偶數(shù),1+2^r-s為奇數(shù).

∴(*)為矛盾等式,不成立.

故這樣的三項(xiàng)不存在.