dannyljl
幼苗
共回答了18个问题采纳率:94.4% 举报
(1)∵点(2S n +a n ,S n+1 )在f(x)=
1
2 x+
1
3 的图象上
∴ S n+1 =
1
2 (2 S n + a n )+
1
3 ,
即 S n+1 - S n =
1
2 a n +
1
3 ,
a n+1 =
1
2 a n +
1
3 ,
即 a n+1 -
2
3 =
1
2 ( a n -
2
3 ) ,
∴ a 1 -
2
3 =
1
2 ≠0 ,
∴数列{ a n -
2
3 }是等比数列.
(2)由(1)知, a n -
2
3 =( a 1 -
2
3 )(
1
2 ) n-1 ,
得 a n =
2
3 + (
1
2 ) n ,
∵
b n+1
b n =
n
n+1 ,
∴
b 2
b 1 =
1
2 ,
b 3
b 2 =
2
3 ,
b 4
b 3 =
3
4 ,…,
b n
b n-1 =
n-1
n ,
∴
b n
b 1 =
1
2 ×
2
3 ×
3
4 ×…×
n-1
n =
1
n ,
即 b n =
1
n b 1 =
1
n (n≥2).
又∵b 1 =1,∴ b n =
1
n .
(3) c n =
a n -
2
3
b n =
(
1
2 ) n
1
n =n• (
1
2 ) n ,
T n =1×
1
2 +2× (
1
2 ) 2 +…+n× (
1
2 ) n ,①
1
2 T n =1× (
1
2 ) 2 +…+ (
1
2 ) n -n• (
1
2 ) n+1 ,②
①-②得:
1
2 T n =
1
2 + (
1
2 ) 2 +…+ (
1
2 ) n -n(
1
2 ) n+1 ,
1
2 T n =
1
2 (1-
1
2 n )
1-
1
2 -n(
1
2 ) n+1 ,
1
2 T n =1-
1
2 n -n(
1
2 ) n+1 ,
T n =2-
2+n
2 n ,
T n -1=1-
2+n
2 n ,
n=1时,T n -1<0,即T n <1,
n=2时,T n -1=0,即T n =1,
n≥3时,T n -1>0,即T n >1.
1年前
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