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

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

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