a1=a
a(n+1)=3(an +1)/(an +3)
a(n+1) + √3 = 3(an +1)/(an +3) + √3
= [(3+√3)an + (3√3+3) ]/(an +3)
=(3+√3)( an +√3) /(an +3)
1/[a(n+1) + √3] = (an + 3)/[(3+√3)( an +√3) ]
= 1/(3+√3) + (3-√3)/(an +√3)
1/[a(n+1) + √3] + 1/(3-√3) = (3-√3) [ 1/(an +√3) + 1/(3-√3)]
=>{1/(an +√3) + 1/(3-√3)} 是等比数列,q=(3-√3)
1/(an +√3) + 1/(3-√3) = (3-√3)^(n-1) .[1/a1 + 1/(3-√3)]
= (3-√3)^(n-1) .[1/a + 1/(3-√3)]
1/(an +√3) = (3-√3)^(n-1)/a + (3-√3)^(n-2) - 1/(3-√3)
an = -√3 + 1/[(3-√3)^(n-1)/a + (3-√3)^(n-2) - 1/(3-√3)]从a(n+1)=3(an +1)/(an +3)这步到a(n+1) + √3 = 3(an +1)/(an +3) + √3这步，如何知道要配根号3，有什么技巧么
顺便问下这题用特征根法能解不，试了下感觉计算量好大
话说你还真是碉堡了，解答都看了好一会才明白a(n+1)=3(an +1)/(an +3)

The aux. equation
x=3(x+1)/(x+3)
x^2+3x=3x+3
x=√3 or -√3

x1=√3, x2= -√3

a(n+1)-x1=3(an +1)/(an +3)- x1or
a(n+1)-x2=3(an +1)/(an +3)-x2

a(n+1) + √3 = 3(an +1)/(an +3) + √3
= [(3+√3)an + (3√3+3) ]/(an +3)
=(3+√3)( an +√3) /(an +3)
1/[a(n+1) + √3] = (an + 3)/[(3+√3)( an +√3) ]
= 1/(3+√3) + (3-√3)/(an +√3)

let
1/[a(n+1) + √3] +k= (3-√3)[1/(an +√3)+k ]
(2-√3)k =1/(3+√3)
k =1/(3-√3)

ie
1/[a(n+1) + √3] = 1/(3+√3) + (3-√3)/(an +√3)
1/[a(n+1) + √3] + 1/(3-√3) = (3-√3) [ 1/(an +√3) + 1/(3-√3)]
=>{1/(an +√3) + 1/(3-√3)} 是等比数列, q=(3-√3)