(2009•烟台二模)已知公差大于零的等差数列an的前n项和为Sn,且满足:a3•a4=117,a2+a5=22.

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  • (1)an为等差数列,a3•a4=117,a2+a5=22

    又a2+a5=a3+a4=22

    ∴a3,a4是方程x2-22x+117=0的两个根,d>0

    ∴a3=9,a4=13

    a1+2d=9

    a1+3d=13

    ∴d=4,a1=1

    ∴an=1+(n-1)×4=4n-3

    (2)由(1)知,sn=n+

    n(n−1)×4

    2=2n2−n

    ∵bn=

    sn

    n+c=

    2n2−n

    c+n

    ∴b1=

    1

    1+c,b2=

    6

    2+c,b3=

    15

    3+c,

    ∵bn是等差数列,∴2b2=b1+b3,∴2c2+c=0,

    ∴c=−

    1

    2(c=0舍去),

    当c=−

    1

    2时,bn=2n为等差数列,满足要求.

    (3)由(2)得bn=

    2n2−n

    n−

    1

    2=2n,Tn=2n+

    n(n−1)×2

    2=n2+n=(n+1)n

    2Tn-3bn-1=2(n2+n)-3(2n-2)=2(n-1)2+4≥4,

    但由于n=1时取等号,从而等号取不到2Tn-3bn-1=2(n2+n)-3(2n-2)=2(n-1)2+4>4,

    64bn

    (n+9)bn+1=

    64×2n

    (n+9)•2(n+1)=

    64n

    n2+10n+9=

    64

    n+

    9

    n+10≤4,

    n=3时取等号(15分)

    (1)、(2)式中等号不能同时取到,所以2Tn−3bn−1>

    64bn

    (n+9)bn+1.