题目

已知数列an满足a1=,anan+1=()n,n∈N*.(1)求数列{an}的通项公式;(2)设a＞0,数列bn满足b1=,bn+1=,若|bn|≤an对n∈N*成立,试求a的取值范围. 答案：解:(1)=,∴=.                                                            又∵a1=,a1a2=·,∴a2=.                                                                       ∴a1,a3,a5,…,a2n-1,…及a2,a4,…,a2n,…均为公比为的等比数列.∵a2n-1=()2n-1,a2n=()2n,∴an=()n.(2) |b1|≤||≤或a≥2.现证:a≥2时,|bn|≤an对n∈N*成立.①n=1时,|b1|≤a1成立;②假设n=k(k≥1)时,|bk|≤ak成立,则n=k+1时,|bk+1|=≤≤≤,                                即n=k+1时,|bk+1|≤ak+1也成立.∴n∈N*时,|bn|≤an.                                                                                            ∴a的取值范围是［2,+∞).

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