已知各项均为正数的数列{a n }的前n项和s n 满足s 1 >1,且6s n =(a n +1)(a n +2)(n为正整数).

2025-04-24 04:09:40
推荐回答(1个)
回答1:

(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 =
3n-1,n为偶数
2 3n-1 ,n为奇数
,T n =b 1 +b 2 +…+b n .…..(7分)
当n为偶数时,
T n =(b 1 +b 3 +…+b n-1 )+(b 2 +b 4 +…+b n
=
4(1- 8
n
2
)
1-8
+
n
2
(5+3n-1)
2
=
4
7
( 8
n
2
-1)+
n(3n+4)
4
,….(9分)
当n为奇数时,T n =(b 1 +b 3 +…+b n )+(b 2 +b 4 +…+b n-1
=
4(1- 8
n+1
2
)
1-8
+
n-1
2
(5+3n-4)
2
=
4
7
( 8
n+1
2
-1)+
(n-1)(3n+1)
4
.…(11分)∴ T n =
4
7
( 8
n
2
-1)+
n(3n-4)
4
,n为偶数
4
7
( 8
n+1
2
-1)+
(n-1)(3n+1)
4
,n为奇数
…..(12分)
(3) C n =
2 a n+1
a n
=
2 3n+2
3n-1
,n为偶数
a n+1
2 a n
=
3n+2
2 3n-1
,n为奇数
,…..(14分)
当n为奇数时, C n+2 - C n =
3n+8
2 3n+5
-
3n+2
2 3n-1
=
1
2 3n+5
[3n+8-64(3n+2)]<0
,…(15分)
∴C n+2 <C n
∴{C n }递减,…..(16分)
C n C 1 =
5
4
<2008
,…..(17分)
因此不存在满足条件的正整数N.…..(18分)