Question

已知數列a_n滿足a₁=,a_na_n+1=()ⁿ,n∈N^*.

(1)求數列{a_n}的通項公式;

(2)設a＞0,數列b_n滿足b₁=,b_n+1=,若|b_n|≤a_n對n∈N^*成立,試求a的取值范圍.

Answer 1

解:(1)=,∴=.                                                            ?

又∵a₁=,a₁a₂=·,∴a₂=.                                                                       ?

∴a₁,a₃,a₅,…,a_2n-1,…及a₂,a₄,…,a_2n,…均為公比為的等比數列.?

∵a_2n-1=()^2n-1,a_2n=()²ⁿ,∴a_n=()ⁿ.?

(2) |b₁|≤||≤或a≥2.

?

現證:a≥2時,|b_n|≤a_n對n∈N^*成立.?

①n=1時,|b₁|≤a₁成立;?

②假設n=k(k≥1)時,|b_k|≤a_k成立,?

則n=k+1時,|b_k+1|=≤≤≤,                                ?

即n=k+1時,|b_k+1|≤a_k+1也成立.?

∴n∈N^*時,|b_n|≤a_n.                                                                                            ?

∴a的取值范圍是［2,+∞).