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

因为a1=2，an+1=2an/（an+2）
所以1/（an+1）=（an+2）/2an=1/塌敬2+1/an
所以1/an+1-1/an=1/2
所以{1/an}为等差数薯弊列
所以1/an=1/a1+（n-1）*1/2
所以1/an=n/2
所以团手慎an=2/n

望采纳哦 亲 O(∩_∩)O~

Answer 2

解：春岩∵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
解逗宏得不动点是扒指御：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 3

由题可得灶简，1/a(n+1)=(1/an)+1/2，所以州辩凳1/册旅an=(n+1)/2，所以an=2/(n+1)