1. Proof: Since $F_{1}=F_{2}=1$, (91) holds for $n=1$. Now assume (91) holds for $1 \leqslant n \leqslant k$, we consider the case $n=k+1$. By equation (8) in this chapter, we have
$$\begin{array}{c} F_{k+1}=F_{k}+F_{k-1} \\ =\frac{(1+\sqrt{5})^{k}-(1-\sqrt{5})^{k}+2(1+\sqrt{5})^{k-1}-2(1-\sqrt{5})^{k-1}}{2^{k} \sqrt{5}} \\ =\frac{(1+\sqrt{5})^{k+1}}{2^{k+1} \sqrt{5}}\left(\frac{2}{1+\sqrt{5}}-\frac{2(1-\sqrt{5})^{k}}{(1+\sqrt{5})^{k+1}}\right. \\ \left.+\frac{4}{(1+\sqrt{5})^{2}}-\frac{4(1-\sqrt{5})^{k-1}}{(1+\sqrt{5})^{k+1}}\right) \\ =\frac{(1+\sqrt{5})^{k+1}}{2^{k+1} \sqrt{5}}\left(1-\frac{(1-\sqrt{5})^{k-1}}{(1+\sqrt{5})^{k+1}}(6-2 \sqrt{5})\right) \\ =\frac{(1+\sqrt{5})^{k+1}}{2^{k+1} \sqrt{5}}\left(1-\frac{(1-\sqrt{5})^{k+1}}{(1+\sqrt{5})^{k+1}}\right) \\ =\frac{(1+\sqrt{5})^{k+1}-(1-\sqrt{5})^{k+1}}{2^{k+1} \sqrt{5}}, \end{array}$$

This proves that (91) holds for $n=k+1$, thus (91) is proved.
Notice that for $n \geqslant 1$ we have
$$\sqrt{5}-1>0$$

and
$$(\sqrt{5}-1)^{n}<2^{n}$$

Thus, we always have
$$0<\frac{(\sqrt{5}-1)^{n}}{\sqrt{5} \cdot 2^{n}}<1$$

Since $F_{n}$ is a positive integer, (92) holds.