Question

(理)已知一列非零向量a _n,n∈N^*,滿足:a₁=(10,-5), a _n=(x_n,y_n)=k(x_n-1-y_n-1,x_n-1+y_n-1)(n≥2),其中k是非零常數(shù).

(1)求數(shù)列{| a _n|}的通項(xiàng)公式;

(2)求向量a _n-1與a _n的夾角(n≥2);

(3)當(dāng)k=時(shí),把a ₁, a ₂,…, a _n,…中所有與a ₁共線的向量按原來的順序排成一列,記為b₁,b₂,…,b_n,…,令OB_n=b₁+b₂+…+b_n,O為坐標(biāo)原點(diǎn),求點(diǎn)列{B_n}的極限點(diǎn)B的坐標(biāo).〔注:若點(diǎn)坐標(biāo)為(t_n,s_n),且t_n=t,s_n=s,則稱點(diǎn)B(t,s)為點(diǎn)列的極限點(diǎn)〕

(文)設(shè)函數(shù)f(x)=5^x-6,g(x)=f(x).

(1)解不等式g(n)［g(1)+g(2)+…+g(n)］＜0(n∈N^*);

(2)求h(n)=g(n)［g(1)+g(2)+…+g(n)］-132n(n∈N^*)的最小值.

Answer 1

答案：(理)解:(1)| a _n|==

==|k||a_n-1|(n≥2),

∴|k|≠0,|a₁|=.

∴{|a_n|}是首項(xiàng)為,公比為|k|的等比數(shù)列.∴|a_n|=(|k|)^n-1.

(2)a_n·a_n-1=k(x_n-1-y_n-1,x_n-1+y_n-1)·(x_n-1,y_n-1)=k(x_n-1²+y_n-1²)=k|a_n-1|²,

∴cos〈a_n,a_n-1〉=

∴當(dāng)k＞0時(shí),〈a,a_n-1〉=,當(dāng)k＜0時(shí),〈a_n,a_n-1〉=.

(3)當(dāng)k=時(shí),由(2)知4〈a_n,a_n-1〉=π,

∴每相隔3個(gè)向量的兩個(gè)向量必共線,且方向相反.

∴與向量a₁共線的向量為{a₁,a₅,a₉,a₁₃,…}={b₁,b₂,b₃,b₄,…}.

記a_n的單位向量為a_n0,則a₁=|a₁|a_n0,

則a_n=|a_n|a_n0=|a₁|(|k|)^n-1a_n0,b_n=a_4n-3=|a₁|(|k|)^4n-4(-1)^n-1a_n0=a₁(-4|k|⁴)^n-1=(10,-5)(-)^n-1.

設(shè)=(t_n,s_n),則t_n=10［1+(-)+(-)²+…+(-)^n-1］=10×,

∴.∴點(diǎn)列{B_n}的極限點(diǎn)B的坐標(biāo)為(8,-4).

(文)解:(1)g(n)=(5^n-6)=2n-12(n∈N^*),

∴g(1)+g(2)+…+g(n)=n²-11n.

解不等式(2n-12)(n²-11n)＜0,得6＜n＜11(n∈N^*).

(2)當(dāng)x∈R時(shí),h(x)=(2x-12)(x²-11x)-132x=2x³-34x²,h′(x)=6x²-68x,

由h′(x)＞0,得x＜0或x＞11,

∵n∈N^*,∴1≤n≤11時(shí),h(n)單調(diào)遞減,n≥12時(shí),h(n)單調(diào)遞增.

當(dāng)n=11時(shí),h(11)=-1 452,當(dāng)n=12時(shí),h(12)=-1 440,∴h(n)_min=h(11)=-1 452.