Question

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

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

(2)这个数列从第几项开始以后各项均小于1/1000

Answer 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

Answer 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