设f(x)=x 2 ,g(x)=8x,数列{a n }(n∈N*)满足a 1 =2,(a n+1 -a n )•g(a

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  • (Ⅰ)由已知,得(a n+1-a n)•8(a n-1)+(a n-1) 2=0.

    即(a n-1)(8a n+1-7a n-1)=0.

    ∵a 1=2≠1,∴a 2≠1,同理a 3≠1,…,a n≠1.

    ∴8a n+1=7a n+1.

    即8(a n+1-1)=7(a n-1),

    ∴数列{a n-1}是以a 1-1=1为首项,

    7

    8 为公比的等比数列.

    (Ⅱ)由(1),得 a n -1=(

    7

    8 ) n-1 .

    ∴ b n =(n+1)•(

    7

    8 ) n .

    则 b n+1 =(n+2)•(

    7

    8 ) n+1 .

    b n+1

    b n =

    n+2

    n+1 •

    7

    8 ,设

    b n+1

    b n ≥1,则n≤6.

    因此,当n<6时,b n<b n+1;当n=6时,b 6=b 7,当n>6时,b n>b n+1

    ∴当n=6或7时,b n取得最大值.

    (Ⅲ) S n =2•

    7

    8 +3•(

    7

    8 ) 2 +4•(

    7

    8 ) 3 +…+n•(

    7

    8 ) n-1 +(n+1)•(

    7

    8 ) n

    7

    8 • S n =2•(

    7

    8 ) 2 +3•(

    7

    8 ) 3 +4•(

    7

    8 ) 4 +…+n•(

    7

    8 ) n +(n+1)•(

    7

    8 ) n+1

    相减得:

    1

    8 • S n =2•

    7

    8 +(

    7

    8 ) 2 +(

    7

    8 ) 3 +…+(

    7

    8 ) n -(n+1)•(

    7

    8 ) n+1 =

    7

    8 +

    7

    8 ×8×[1-(

    7

    8 ) n ]-(n+1)•(

    7

    8 ) n+1

    =

    63

    8 -(n+9)•(

    7

    8 ) n+1

    ∴ S n =63-8(n+9)•(

    7

    8 ) n+1 .