已知各项均为正数的数列{a n }的前n项和为S n ,且S n ,a n , 1 2 成等差数列,

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  • (1)由S n,a n

    1

    2 成等差数列,可得 2 a n = S n +

    1

    2 ,∴ a 1 =

    1

    2 ,a 2=1

    (2)由 2 a n = S n +

    1

    2 可得,2S n=4a n-1(n≥1),∴2S n-1=4a n-1-1(n≥2)

    ∴两式相减得2a n=(4a n-1)-(4a n-1-1)=4a n-4a n-1,即a n=2a n-1(n≥2),

    ∴数列{a n}是以

    1

    2 为首项,以2为公比的等比数列,

    ∴ a n =

    1

    2 × 2 n-1 = 2 n-2 (n∈N *

    (3)由题意可得, C n =(4-2n)×(

    1

    2 ) n-2

    T n=C 1+C 2+…+C n

    = 2×(

    1

    2 ) -1 +0×(

    1

    2 ) 0 +(-2)×(

    1

    2 ) 1 +…+(4-2n)× (

    1

    2 ) n-2

    1

    2 T n =2×(

    1

    2 ) 0 +0×(

    1

    2 ) 1 +…+(4-2n)× (

    1

    2 ) n-1

    错位相减可得,

    1

    2 T n =2n×(

    1

    2 ) n-1

    T n =4n×(

    1

    2 ) n-1