(2010•天津模拟)已知数列{an}的前n项和Sn满足:Sn=a(Sn-an+1)(a为常数,a≠0,a≠1).

1个回答

  • 解题思路:(Ⅰ)由题意知a1=a,Sn=a(Sn-an+1),Sn-1=a(Sn-1-an-1+1),由此可知an=a•an-1

    a

    n

    a

    n-1

    =a

    ,所以an=a•an-1=an

    (Ⅱ)由题意知a≠1,

    b

    n

    =(

    a

    n

    )

    2

    +

    a(

    a

    n

    -1)

    a-1

    a

    n

    b

    n

    =

    (2a-1)

    a

    2n

    -a

    a

    n

    a-1

    ,由此可解得

    a=

    1

    2

    (Ⅲ)证明:由题意知

    b

    n

    =(

    1

    2

    )

    n

    ,所以

    c

    n

    =

    1

    (

    1

    2

    )

    n

    +1

    -

    1

    (

    1

    2

    )

    n+1

    -1

    =

    2-

    1

    2

    n

    +1

    +

    1

    2

    n+1

    -1

    ,由此可知Tn>2n-[1/2].

    (Ⅰ)S1=a(S1-a1+1)

    ∴a1=a,.(1分)

    当n≥2时,Sn=a(Sn-an+1),Sn-1=a(Sn-1-an-1+1),

    两式相减得:an=a•an-1

    an

    an−1=a

    (a≠0,n≥2)即{an}是等比数列.

    ∴an=a•an-1=an;(4分)

    (Ⅱ)由(Ⅰ)知a≠1,

    bn=(an)2+

    a(an−1)

    a−1an,bn=

    (2a−1)a2n−aan

    a−1,

    若{bn}为等比数列,则有b22=b1b3

    而b1=2a2,b2=a3(2a+1),b3=a4(2a2+a+1)(6分)

    故[a3(2a+1)]2=2a2•a4(2a2+a+1),解得a=

    1

    2,(7分)

    再将a=[1/2]代入得bn=([1/2])n成立,所以a=[1/2].(8分)

    (Ⅲ)证明:由(Ⅱ)知bn=(

    1

    2)n,

    所以cn=

    1

    (

    1

    2)n+1−

    1

    (

    1

    2)n+1−1=

    2n

    2n+1+

    2n+1

    2n+1−1=2−

    1

    2n+1+

    1

    2n+1−1(10分)

    所以cn>2−

    1

    2n+

    1

    2n+1

    Tn=c1+c2++cn>(2−

    点评:

    本题考点: 数列的应用;数列的求和;数列递推式.

    考点点评: 本题考查数列知识的综合应用,解题时要认真审题,仔细解答.