(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 n③
又2T n=1×2 2+2×2 3+…+(n-1)•2 n+n2 n+1 ④
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