設(shè)數(shù)列{
bn}滿足
bn+2=-
bn+1-
bn(
n∈N
*),
b2=2
b1.
(1)若
b3=3,求
b1的值;
(2)求證數(shù)列{
bnbn+1bn+2+
n}是等差數(shù)列;
(3)設(shè)數(shù)列{
Tn}滿足:
Tn+1=
Tnbn+1(
n∈N
*),且
T1=
b1=-

,若存在實數(shù)
p,
q,對任意
n∈N
*都有
p≤
T1+
T2+
T3+…+
Tn<
q成立,試求
q-
p的最小值.
(1)
b1=-1(2)見解析(3)

(1)∵
bn+2=-
bn+1-
bn,
∴
b3=-
b2-
b1=-3
b1=3,
∴
b1=-1;(3分)
(2)∵
bn+2=-
bn+1-
bn①,
∴
bn+3=-
bn+2-
bn+1②,
②-①得
bn+3=
bn,(5分)
∴(
bn+1bn+2bn+3+
n+1)-(
bnbn+1bn+2+
n)=
bn+1bn+2(
bn+3-
bn)+1=1為常數(shù),
∴數(shù)列{
bnbn+1bn+2+
n}是等差數(shù)列.(7分)
(3)∵
Tn+1=
Tn·
bn+1=
Tn-1bnbn+1=
Tn-2bn-1bnbn+1=…=
b1b2b3…
bn+1當
n≥2時
Tn=
b1b2b2…
bn(*),
當
n=1時,
T1=
b1適合(*)式
∴
Tn=
b1b2b3…
bn(
n∈N
*).(9分)
∵
b1=-

,
b2=2
b1=-1,
b3=-3
b1=

,
bn+3=
bn,
∴
T1=
b1=-

,
T2=
T1b2=

,
T3=
T2b3=

,
T4=
T3b4=
T3b1=
T1,
T5=
T4b5=
T2b3b4b5=
T2b1b2b3=
T2,
T6=
T5b6=
T3b4b5b6=
T3b1b2b3=
T3,
……
T3n+1+
T3n+2+
T3n+3=
T3n-2b3n-1b3nb3n+1+
T3n-1b3nb3n+1b3n+2+
T3nb3n+1b3n+2b3n+3=
T3n-2b1b2b3+
T3n-1b1b2b3+
T3nb1b2b3=

(
T3n-2+
T3n-1+
T3n),
∴數(shù)列{
T3n-2+
T3n-1+
T3n)(
n∈N
*)是等比數(shù)列,
首項
T1+
T2+
T3=

且公比
q=

,(11分)記
Sn=
T1+
T2+
T3+…+
Tn,
①當
n=3
k(
k∈N
*)時,
Sn=(
T1+
T2+
T3)+(
T4+
T5+
T6)…+(
T3k-2+
T3k-1+
T3k)
=

,
∴

≤
Sn<3;(13分)
②當
n=3
k-1(
k∈N
*)時
Sn=(
T1+
T2+
T3)+(
T4+
T5+
T6)+…+(
T3k-2+
T3k-1+
T3k)-
T3k=3

-(
b1b2b3)
k=3-4·

∴0≤
Sn<3;(14分)
③當
n=3
k-2(
k∈N
*)時
Sn=(
T1+
T2+
T3)+(
T4+
T5+
T6)+…+(
T3k-2+
T3k-1+
T3k)-
T3k-1-
T3k=3

-(
b1b2b3)
k-1b1b2-(
b1b2b3)
k=3

-

k-1-
k=3-

·
k,
∴-

≤
Sn<3.(15分)
綜上得-

≤
Sn<3則
p≤-

且
q≥3,∴
q-
p的最小值為

.(16分)
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