(Ⅰ)∵bn+2=-bn+1-bn, ∴b3=-b2-b1=-3b1=3, ∴b1=-1.(3分) (Ⅱ)∵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为常数 ∴数列{bnbn+1bn+2+n}是等差数列. (7分) (Ⅲ)∵Tn+1=Tn•bn+1=Tn-1bnbn+1=Tn-2bn-1bnbn+1=…=b1b2b3…bn+1 当n≥2时Tn=b1b2b3…bn(*),当n=1时T1=b1适合(*)式 ∴Tn=b1b2b3…bn(n∈N*). (9分) ∵b1=-,b2=2b1=-1,b3=-3b1=,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), ∴数列{T3n-2+T3n-1+T3n}(n∈N*)是等比数列 首项T1+T2+T3=且公比q= (11分) 记Sn=T1+T2+T3+…+Tn ①当n=3k(k∈N*)时,Sn=(T1+T2+T3)+(T4+T5+T6)…+(T3k-2+T3k-1+T3k)==3[1-()k] ∴≤Sn<3; (13分) ②当n=3k-1(k∈N*)时Sn=(T1+T2+T3)+(T4+T5+T6)…+(T3k-2+T3k-1+T3k)-T3k =3[1-()k]-(b1b2b3)k=3-4•()k ∴0≤Sn<3; (14分) ③当n=3k-2(k∈N*)时Sn=(T1+T2+T3)+(T4+T5+T6)…+(T3k-2+T3k-1+T3k)-T3k-1-T3k =3[1-()k]-(b1b2b3)k-1b1b2-(b1b2b3)k=3[1-()k]-()k-1-()k=3-•()k ∴-≤Sn<3 (15分) 综上得-≤Sn<3则p≤-且q≥3, ∴q-p的最小值为. (16分) |