已知数列(An)满足:A1=1,2^(n-1)*An=A下标(n-1) (n为正整数),n≥2 (1)求数列(An)的通项公式

2024-12-18 06:13:38
推荐回答(2个)
回答1:

(1)2^(n-1)*An=A(n-1),则A(n-1)/An=2^(n-1),
于是(A1/A2)*(A2/A3)*...*[A(n-1)/An]=A1/An=2*(2^2)*...*[2^(n-1)]=2^[1+2+...+(n-1)]
=2^{[1+(n-1)]*n/2}=2^[(n^2)/2],
因为A1=1,代入上式,An=2^[-(n^2)/2]
(2)有An=1/{2^[(n^2)/2]},则使An<1/1000,应有M=2^[(n^2)/2]>1000,n=4时,M=256,n=5时,M>2^10
于是,自A5项开始,以后各项均小于1/1000

回答2:

(1)A_n / A_{n-1} = 2 ^ (-(n-1)), A_n = [A_n / A_{n-1}] * [A_{n-1} / A_{n-2}] * ... * [A_2 / A_1] * A_1 = 2 ^ {n(n-1)/2}
(2)由(1)得:n = 5