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(1)法一:由S n =(
a n +1
2 ) 2 得: 4 S n =
a 2n +2 a n +1 ①, 4 S n+1 =
a 2n+1 +2 a n+1 +1 ②,
②-①得 4 a n+1 =
a 2n+1 -
a 2n +2 a n+1 -2 a n ,得到2(a n+1 +a n )=(a n+1 +a n )(a n+1 -a n )
由题知a n+1 +a n ≠0得a n+1 -a n =2,
又 S 1 = a 1 =(
a 1 +1
2 ) 2 ,化为 4 a 1 =
a 21 +2 a 1 +1 ,解得a 1 =1.
∴数列{a n }是以1为首项,2为公差的等差数列,∴a n =1+(n-1)×1=2n-1,
因此前n项和S n =
n(1+2n-1)
2 =n 2 ;
法二:由 S 1 = a 1 =(
a 1 +1
2 ) 2 ,化为 4 a 1 =
a 21 +2 a 1 +1 ,解得a 1 =1.
当n≥2时, 2
S n = a n +1= S n - S n-1 +1 ,
得到 (
S n -1 ) 2 = S n 1 ,即
S n -
S n-1 =1
所以数列{
S n }是以1为首项,1为公差的等差数列,
∴
S n =1+(n-1)×1 =n,得到 S n = n 2 .
(2)①由b n +2n-1+λ得到其前n项和T n =n 2 +λn,
由题意T n 最小值为T 6 ,即T n ≥T 6 ,n 2 +λn≥36+6λ,
化为
11
2 ≤-
λ
2 ≤
13
2 ,∴λ∈[-13,-11].
②因{bn}是“封闭数列”,设b p +b q =b m (p,q,m∈Z * ,且任意两个不相等 )得
2p-1+λ+2q-1+λ=2m-1+λ,化为λ=2(m-p-q)+1,则λ为奇数.
由任意n∈N * ,都有T n ≠0,且
1
12 <
1
T 1 +
1
T 2 +
1
T 3 +…+
1
T n <
11
18 .
得
1
12 < T 1 <
11
18 ,化为
7
11 <λ<11 ,即λ的可能值为1,3,5,7,9,
又 T n = n 2 +λn >0,因为
1
n(n+λ) =
1
λ (
1
n -
1
n+λ ) ,
检验得满足条件的λ=3,5,7,9,
即存在这样的“封闭数列”{b n },使得对任意n∈N * ,都有T n ≠0,
且
1
12 <
1
T 1 +
1
T 2 +
1
T 3 +…+
1
T n <
11
18 .,
所以实数λ的所有取值集合为{3,5,7,9}.
1年前
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