Question

在平面直角坐标系中,已知A_n(n,a_n)、B_n(n,b_n)、C_n(n-1,0)(n∈N^*),满足向量与向量共线,且点B_n(n,b_n)(n∈N^*)都在斜率为6的同一条直线上.

(1)试用a₁,b₁与n来表示a_n;

(2)设a₁=a,b₁=-a,且12＜a≤15,求数列{a_n}中的最小项.

Answer 1

解:(1)∵点B_n(n,b_n)(n∈N^*)都在斜率为6的同一条直线上,?

∴=6,即b_n+1-b_n=6.?

于是数列{b_n}是等差数列,故b_n=b₁+6(n-1).                                                              

∵=(1,a_n+1-a_n),=(-1,-b_n),又与共线,?

∴1×(-b_n)-(-1)(a_n+1-a_n)=0,?

即a_n+1-a_n=b_n.                                                                                                          ?

∴当n≥2时,a_n=a₁+(a₂-a₁)+(a₃-a₂)+…+(a_n-a_n-1)=a₁+b₁+b₂+b₃+…+b_n-1=a₁+b₁(n-1)+3(n-1)

(n-2).                                                                                                                      ?

当n=1时,上式也成立.?

∴a_n=a₁+b₁(n-1)+3(n-1)(n-2).                                                                                   ?

(2)把a₁=a,b₁=-a代入上式,得a_n=a-a(n-1)+3(n-1)(n-2)=3n²-(9+a)n+6+2a.?

∵12＜a≤15,∴＜≤4.?

∴当n=4时,a_n取最小值,最小值为a₄=18-2a.