各项均为正数的数列{a n }中,a 1 =1,S n 是数列{a n }的前n项和,对任意n∈N * ,有2S n =

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  • (1)∵a 1=1,对任意的n∈N*,有2S n=2pa n 2+pa n-p

    ∴2a 1=2pa 1 2+pa 1-p,即2=2p+p-p,解得p=1;

    (2)2S n=2a n 2+a n-1,①

    2S n-1=2a n-1 2+a n-1-1,(n≥2),②

    ①-②即得(a n-a n-1-

    1

    2 )(a n+a n-1)=0,

    因为a n+a n-1≠0,所以a n-a n-1-

    1

    2 =0,

    ∴ a n =

    n+1

    2

    (3)2S n=2a n 2+a n-1=2×

    (n+1) 2

    4 +

    n+1

    2 -1 ,

    ∴S n=

    n 2 +3n

    4 ,

    ∴ b n =

    4 S n

    n+3 • 2 n =n•2 n

    T n=1×2 1+2×2 2+…+n•2 n

    又2T n=1×2 2+2×2 3+…+(n-1)•2 n+n2 n+1

    ④-③T n=-1×2 1-(2 2+2 3+…+2 n)+n2 n+1=(n-1)2 n+1+2

    ∴T n=(n-1)2 n+1+2