Question

(理)數(shù)列{a_n},{b_n}(n=1,2,3,…)由下列條件所確定:

(1)a₁＜0,b₁＞0;

(2)k≥2時,a_k與b_k滿足如下條件:

當(dāng)a_k-1+b_k-1≥0時,a_k=a_k-1,b_k=;

當(dāng)a_k-1+b_k-1＜0時,a_k=,b_k=b_k-1.

那么,當(dāng)a₁=-5,b₁=5時,{a_n}的通項公式a_n=當(dāng)b₁＞b₂＞…＞b_n(n≥2)時,且a₁、b₁表示{b_k}的通項b_k=_______________(k=2,3,…,n).

Answer 1

答案：(理)-5()^n-2  a₁+(b₁-a₁)()^k-1

①a₁+b₁=0,∴a₂=-5,b₂=0,a₂+b₂=-5＜0,a₃=-,b₃=0,a₃+b₃=-＜0,a₃=-,b₄=0.

故n≥2時,a_n=-5()^n-2.

②∵b₁＞b₂＞b₃＞…＞b_n(n≥2),∴a_k-1+b_k-1≥0恒成立.∴{a_n}為常數(shù)列,a_n=a₁,

∴2b_k=a₁+b_k-1.∴2(b_k-a₁)=b_k-1-a₁.

∴數(shù)列{b_n-a₁}是以b₁-a₁為首項,為公比的等比數(shù)列.∴a_k=a₁+(b₁-a₁)()^k-1.