$ 解：当n=1时，原式=\frac 1{\sqrt{5}}[(\frac {1+\sqrt{5}}2)^1-(\frac {1-\sqrt {5}}2)^1]$
$ =\frac 1{\sqrt{5}}×[\frac {1+\sqrt{5}-1+\sqrt{5}}2]$
$ =\frac 1{\sqrt{5}}×\sqrt{5}$
$ =1$
$ ∴第1个数为1$
$ 当n=2时，原式=\frac 1{\sqrt{5}}[(\frac {1+\sqrt{5}}2)^2-(\frac {1-\sqrt {5}}2)^2$
$ =\frac 1{\sqrt{5}}×[\frac {1+2\sqrt{5}+5}4-\frac {1-2\sqrt{5}+5}4]$
$ =\frac 1{\sqrt{5}}×\frac {6+2\sqrt{5}-6+2\sqrt{5}}4$
$ =\frac 1{\sqrt{5}}×\sqrt{5}$
$=1$
$∴第2个数为1$
$(2)x_{n+1}-x_n=\frac 1{\sqrt {5}}[(\frac {1+\sqrt {5}}2)^{n+1}-(\frac {1-\sqrt {5}}2)^{n+1}]-\frac 1{\sqrt {5}}[(\frac {1+\sqrt {5}}2)^{n}-(\frac {1-\sqrt {5}}2)^{n}]$
$=\frac 1{\sqrt {5}}×[(\frac {1+\sqrt {5}}2)^{n+1}-(\frac {1+\sqrt {5}}2)^n-(\frac {1-\sqrt {5}}2)^{n+1}+(\frac {1-\sqrt {5}}2)^n]$
$=\frac 1{\sqrt {5}}×[(\frac {1+\sqrt {5}}2)^n×(\frac {1+\sqrt {5}}2-1)-(\frac {1-\sqrt {5}}2)^n×(\frac {1-\sqrt {5}}2-1)]$
$=\frac 1{\sqrt {5}}×[(\frac {1+\sqrt {5}}2)^{n-1}×\frac {1+\sqrt {5}}2×\frac {\sqrt {5}-1}2-(\frac {1-\sqrt {5}}2)^{n-1}×\frac {1-\sqrt {5}}2×\frac {-1-\sqrt {5}}2]$
$=\frac 1{\sqrt {5}}×[(\frac {1+\sqrt {5}}2)^{n-1}-(\frac {1-\sqrt {5}}2)^{n-1}]$
$=x_{n-1}$
$即x_{n+1}-x_n=x_{n-1}$
