(1)n=1时,6a 1 =a 1 2 +3a 1 +2,且a 1 >1,解得a 1 =2.…..(2分) n≥2时,6S n =a n 2 +3a n +2,6S n-1 =a n-1 2 +3a n-1 +2, 两式相减得:6a n =a n 2 -a n-1 2 +3a n -3a n-1 即(a n +a n-1 )(a n -a n-1 -3)=0, ∵a n +a n-1 >0, ∴a n -a n-1 =3, ∴{a n }为等差数列,a n =3n-1.….(6分) (2) b n =
当n为偶数时, T n =(b 1 +b 3 +…+b n-1 )+(b 2 +b 4 +…+b n ) =
当n为奇数时,T n =(b 1 +b 3 +…+b n )+(b 2 +b 4 +…+b n-1 ) =
(3) C n =
当n为奇数时, C n+2 - C n =
∴C n+2 <C n , ∴{C n }递减,…..(16分) C n ≤ C 1 =
因此不存在满足条件的正整数N.…..(18分) |