Question

已知数列{An}满足A1=1，An+1=2An/An+2，求数列{An}的通项公式？

Answer 1

a(n+1)=2an/(an+2)
1/a(n+1)=(an+2)/(2an)=1/an+1/2
1/a(n+1)-1/an=1/2，为定值。
1/a1=1/1=1
数列{1/an}是以1为首项，1/2为公差的等差数列。
1/an=1/a1+(n-1)(1/2)=1+(n-1)/2=(n+1)/2
an=2/(n+1)
n=1时，a1=2/(1+1)=1，同样满足。
数列{an}的通项公式为an=2/(n+1)。

Answer 2

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]