Question

已知数列a_n满足a₁=,a_na_n+1=()ⁿ,n∈N^*.

(1)求数列{a_n}的通项公式;

(2)设a＞0,数列b_n满足b₁=,b_n+1=,若|b_n|≤a_n对n∈N^*成立,试求a的取值范围.

Answer 1

解:(1)=,∴=.                                                            ?

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

∴a₁,a₃,a₅,…,a_2n-1,…及a₂,a₄,…,a_2n,…均为公比为的等比数列.?

∵a_2n-1=()^2n-1,a_2n=()²ⁿ,∴a_n=()ⁿ.?

(2) |b₁|≤||≤或a≥2.

?

现证:a≥2时,|b_n|≤a_n对n∈N^*成立.?

①n=1时,|b₁|≤a₁成立;?

②假设n=k(k≥1)时,|b_k|≤a_k成立,?

则n=k+1时,|b_k+1|=≤≤≤,                                ?

即n=k+1时,|b_k+1|≤a_k+1也成立.?

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

∴a的取值范围是［2,+∞).