Question

已知数列[an]满足a1=1，an+1=an+n（n是整数）求数列[an]的通项公式

Answer 1

解：∵数列{a[n]}满足a[n
1]=(a[n]
2)/(a[n]
1)
采用不动点法，设：x=(x
2)/(x
1)
x^2=2
解得不动点是：x=±√2
∴(a[n
1]-√2)/(a[n
1]
√2)
={(a[n]
2)/(a[n]
1)-√2}/{(a[n]
2)/(a[n]
1)
√2}
={(a[n]
2)-√2(a[n]
1)}/{(a[n]
2)
√2(a[n]
1)}
={(1-√2)a[n]-(√2-2)}/{(1
√2)a[n]
(√2
2)}
={(1-√2)(a[n]-√2)}/{(1
√2)(a[n]
√2)}
={(1-√2)/(1
√2)}{(a[n]-√2)/(a[n]
√2)}
=(2√2-3){(a[n]-√2)/(a[n]
√2)}
∵a[1]=1
∴(a[1]-√2)/(a[1]
√2)=2√2-3
∴{(a[n]-√2)/(a[n]
√2)}是首项和公比均为2√2-3的等差数列
即：(a[n]-√2)/(a[n]
√2)=(2√2-3)(2√2-3)^(n-1)=(2√2-3)^n
a[n]-√2=a[n](2√2-3)^n
√2(2√2-3)^n
a[n][1-(2√2-3)^n]=√2[1
(2√2-3)^n]
∴{a[n]}的通项公式：a[n]=√2[1
(2√2-3)^n]/[1-(2√2-3)^n]

Answer 2

(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