数列高手请进设X(1)=1,X(n+1)=1+Xn/( X(n)+1),求X(n)的通项公式.

∵x[n+1]=1+x[n]/(x[n]+1)=(2x[n]+1)/(x[n]+1)
∴不动点法,有：y=(2y+1)/(y+1),即：y^2-y-1=0
∴y=(1±√5)/2
∴[x[n+1]-(1+√5)/2]/[x[n+1]-(1-√5)/2]
=[(2x[n]+1)/(x[n]+1)-(1+√5)/2]/[(2x[n]+1)/(x[n]+1)-(1-√5)/2]
=[2(2x[n]+1)-(1+√5)(x[n]+1)]/[2(2x[n]+1)-(1-√5)(x[n]+1)]
=[4x[n]+2-x[n]-√5x[n]-1-√5]/[4x[n]+2-x[n]+√5x[n]-1+√5]
=[(3-√5)x[n]+1-√5]/[(3+√5)x[n]+1+√5]
=[(3-√5)/(3+√5)][x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=[(1-√5)/(1+√5)]^2[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
∵x[1]=1
∴[x[1]-(1+√5)/2]/[x[1]-(1-√5)/2]=(1-√5)/(1+√5)
∴{[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]}是：
首项为(1-√5)/(1+√5),公比为[(1-√5)/(1+√5)]^2的等比数列
即：[x[n]-(1+√5)/2]/[x[n]-(1-√5)/2]
=(1-√5)/(1+√5)[(1-√5)/(1+√5)]^[2(n-1)]
=[(1-√5)/(1+√5)]^(2n-1)
于是有：
x[n]-(1+√5)/2=x[n][(1-√5)/(1+√5)]^(2n-1)-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n]-x[n][(1-√5)/(1+√5)]^(2n-1)=(1+√5)/2-[(1-√5)/2][(1-√5)/(1+√5)]^(2n-1)
x[n][(1+√5)^(2n-1)-(1-√5)^(2n-1)]=[(1+√5)^(2n)-(1-√5)^(2n)]/2
∴{x[n]}的通项公式：
x[n]=(1/2)[(1+√5)^(2n)-(1-√5)^(2n)]/[(1+√5)^(2n-1)-(1-√5)^(2n-1)]

果然有够难的
X(n+1)=1+Xn/( X(n)+1)
X(n+1) = 2 - 1/[X(n)+1]
X(2) = 3/2
X(n+1)+1 = 3 -1/[X(n)+1]
令Z(n) = X(n)+1
Z(1) = 2 Z(2) = 5/2
Z(n+1) = 3 - 1/Z(n)
令Z(n) = P(n)/Q(n)
...

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