(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