一道数列求通项问题

题目

一道数列求通项问题
a1=2
a（n+1）-an=2/【a（n+1）+an-1】,求an

答案

a(n+1) - a(n) = [a(n+1) - 1/2] - [a(n) - 1/2] = 2/[a(n+1) + a(n) - 1] = 2/{[a(n+1)-1/2] + [a(n)-1/2]},
[a(n+1)-1/2]^2 - [a(n)-1/2]^2 = 2,
{ [a(n)-1/2]^2 }是首项为 [a(1)-1/2]^2 = 9/4, 公差为2的等差数列.
[a(n) - 1/2]^2 = 9/4 + 2(n-1) = 2n + 1/4 = (8n+1)/4,
|a(n) - 1/2| = (1/2)(8n+1)^(1/2).
a(1)=2,
n>=2时,
a(n) = [ 1 + (8n+1)^(1/2)]/2
或
a(n) = [1 - (8n+1)^(1/2)]/2