Question

在平面直角坐標(biāo)系中,已知A_n(n,a_n)、B_n(n,b_n)、C_n(n-1,0)(n∈N^*),滿足向量與向量共線,且點(diǎn)B_n(n,b_n)(n∈N^*)都在斜率為6的同一條直線上.

(1)試用a₁,b₁與n來表示a_n;

(2)設(shè)a₁=a,b₁=-a,且12＜a≤15,求數(shù)列{a_n}中的最小項(xiàng).

Answer 1

解:(1)∵點(diǎn)B_n(n,b_n)(n∈N^*)都在斜率為6的同一條直線上,?

∴=6,即b_n+1-b_n=6.?

于是數(shù)列{b_n}是等差數(shù)列,故b_n=b₁+6(n-1).                                                              

∵=(1,a_n+1-a_n),=(-1,-b_n),又與共線,?

∴1×(-b_n)-(-1)(a_n+1-a_n)=0,?

即a_n+1-a_n=b_n.                                                                                                          ?

∴當(dāng)n≥2時(shí),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).                                                                                                                      ?

當(dāng)n=1時(shí),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.?

∴當(dāng)n=4時(shí),a_n取最小值,最小值為a₄=18-2a.