(2014•滨州二模)在已知数列{an}中,a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,其中n为正

1个回答

  • 解题思路:(Ⅰ)由已知条件得

    a

    n+1

    +1=(

    a

    n

    +1

    )

    2

    ,两边取对数,得lg(an+1+1)=2lg(an+1),由此能证明数列{lg(an+1)}是以1为首项,以2为公比的等比数列.

    (Ⅱ)由(Ⅰ)知

    lg(a

    n

    +1)=

    2

    n−1

    ,从而lgTn=lg(a1+1)+lg(a2+1)+…+lg(an+1),由此求出lgTn=2n-1.

    (Ⅲ)由Cn=

    2

    n

    (

    2

    n

    −1)(

    2

    n+1

    −1)

    =

    1

    2

    n

    −1

    1

    2

    n+1

    −1

    ,利用裂项求和法能求出数列{Cn}的前n项和.

    (Ⅰ)证明:∵数列{an}中,a1=9,点(an,an+1)在函数f(x)=x2+2x的图象上,

    ∴an+1=an2+2an,∴an+1+1=(an+1)2,

    对an+1+1=(an+1)2两边取对数,得lg(an+1+1)=2lg(an+1),

    ∵a1=9,∴lg(a1+1)=lg10=1,

    ∴数列{lg(an+1)}是以1为首项,以2为公比的等比数列.

    (Ⅱ)由(Ⅰ)知lg(an+1)=2n−1,

    Tn=(a1+1)…(an+1),

    ∴lgTn=lg(a1+1)(a2+1)…(an+1)

    =lg(a1+1)+lg(a2+1)+…+lg(an+1)

    =20+21+22+…+2n-1

    =

    1−2n

    1−2=2n-1.

    (Ⅲ)证明:Cn=

    lgTn+1

    [lg(an+1+1)−1][lg(an+2+1)−1]

    =

    2n

    (2n−1)(2n+1−1)=[1

    2n−1−

    1

    2n+1−1,

    Sn=(

    1/2−1−

    1

    22−1])+(

    1

    22−1−

    1

    23−1)+(

    1

    23−1−

    1

    24−1)+…+(

    1

    2n−1−

    1

    2n+1−1)

    =1-

    1

    2n+1−1<1.

    ∴Sn<1.

    点评:

    本题考点: 数列与函数的综合;数列的求和.

    考点点评: 本题考查等比数列的证明,考查数列的前n项和的求法,考查不等式的证明,解题时要认真审题,注意裂项求和法的合理运用.