(2009•四川)设数列{an}的前n项和为Sn,对任意的正整数n,都有an=5Sn+1成立,记bn=4+an1−an(

1个回答

  • 解题思路:(Ⅰ)由题设条件能导出an+1-an=5an+1,即

    a

    n+1

    =-

    1

    4

    a

    n

    ,所以

    a

    n

    =(-

    1

    4

    )

    n

    ,∴

    b

    n

    =

    4+

    (-

    1

    4

    )

    n

    1-

    (-

    1

    4

    )

    n

    (Ⅱ)由

    b

    n

    =4+

    5

    (-4)

    n

    -1

    ,知

    c

    n

    =

    b

    2n

    -

    b

    2n-1

    =

    5

    4

    2n

    -1

    +

    5

    4

    2n-1

    +1

    =

    25×

    16

    n

    (

    16

    n

    -1)(

    16

    n

    +4)

    =

    25×

    16

    n

    (

    16

    n

    )

    2

    +3×

    16

    n

    -4

    25×

    16

    n

    (

    16

    n

    )

    2

    =

    25

    16

    n

    ,当n=1时,

    T

    1

    3

    2

    ;当n≥2时,

    T

    n

    4

    3

    +25×(

    1

    16

    2

    +

    1

    16

    3

    +…+

    1

    16

    n

    )

    4

    3

    +25×

    1

    16

    2

    1-

    1

    16

    =

    69

    48

    3

    2

    (Ⅲ)由

    b

    n

    =4+

    5

    (-4)

    n

    -1

    知Rn=b1+b2+…+b2k+1=

    4n+5×(-

    1

    4

    1

    +1

    +

    1

    4

    2

    -1

    -

    1

    4

    3

    +1

    +…-

    1

    4

    2k+1

    +1

    )

    =

    4n+5×[-

    1

    4

    1

    +1

    +(

    1

    4

    2

    -1

    -

    1

    4

    3

    +1

    )+…+(

    1

    4

    2k

    -1

    -

    1

    4

    2k+1

    +1

    )]

    >4n-1.由此入手能推导出正实数λ的最小值为4.

    (Ⅰ)当n=1时,a1=5a1+1,∴a1=−

    1

    4

    又∵an=5Sn+1,an+1=5Sn+1+1

    ∴an+1-an=5an+1,即an+1=−

    1

    4an

    ∴数列an成等比数列,其首项a1=−

    1

    4,公比是q=−

    1

    4

    ∴an=(−

    1

    4)n

    ∴bn=

    4+(−

    1

    4)n

    1−(−

    1

    4)n

    (Ⅱ)由(Ⅰ)知bn=4+

    5

    (−4)n−1

    ∴cn=b2n−b2n−1=

    5

    42n−1+

    5

    42n−1+1=

    25×16n

    (16n−1)(16n+4)

    =

    25×16n

    (16n)2+3×16n−4<

    25×16n

    (16n)2=

    25

    16n

    又b1=3,b2=

    13

    3,∴c1=

    4

    3

    当n=1时,T1<

    3

    2

    当n≥2时,Tn<

    4

    3+25×(

    点评:

    本题考点: 数列递推式;数列的求和;数列与不等式的综合.

    考点点评: 本题主要考查数列、不等式等基础知识、考查化归思想、分类整合思想,以及推理论证、分析与解决问题的能力.