フィボナッチ数のn乗の母関数

(初投稿でMathlogもTexも慣れてないので構文エラーとか変な書き方とかになってたら教えて下さい)

$\displaystyle{\sum_{m=0}^∞(F_m)^nx^m=\frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-x\varphi^{n-2k}}}{\sqrt5^n}}$

フィボナッチ数のn乗の母関数($\displaystyle{\sum_{m=0}^∞(F_m)^nx^m}$)を$\displaystyle{f_n(x)}$と表すと、以下の漸化式が成り立つ。

$\displaystyle{ f_0(x)=\frac{1}{1-x} f_{n+1}(x)=\frac{f_n(x\varphi)-f_n(-\frac{x}{\varphi})}{\sqrt5} }$
冒頭の式の右辺(以下$\displaystyle{g_n(x)}$と表す)がこの漸化式を満たすことを証明する。

証明:
$\displaystyle{\frac{g_n(x\varphi)-g_n(-\frac{x}{\varphi})}{\sqrt5}=\frac{\frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-(x\varphi)\varphi^{n-2k}}}{\sqrt5^n}-\frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-(\frac{-x}{\varphi})\varphi^{n-2k}}}{\sqrt5^n}}{\sqrt5}= \frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-x\varphi^{n-2k+1}}-\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k+x\varphi^{n-2k-1}}}{\sqrt5^{n+1}}= \frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-x\varphi^{n-2k+1}}+\sum_{k=0}^n\frac{{}_nC_{(k+1)-1}}{(-1)^{k+1}-x\varphi^{n-2(k+1)+1}}}{\sqrt5^{n+1}}= \frac{\sum_{k=0}^n\frac{{}_nC_k}{(-1)^k-x\varphi^{n-2k+1}}+\sum_{k=1}^{n+1}\frac{{}_nC_{k-1}}{(-1)^k-x\varphi^{n-2k+1}}}{\sqrt5^{n+1}}= \frac{\frac{1}{1-x\varphi^{n+1}}+\sum_{k=1}^n\frac{{}_nC_k+{}_nC_{k-1}}{(-1)^k-x\varphi^{n-2k+1}}+\frac{1}{-1^{n+1}-x\varphi^{-n-1}}}{\sqrt5^{n+1}}= \frac{\frac{1}{1-x\varphi^{n+1}}+\sum_{k=1}^n\frac{{}_{n+1}C_k}{(-1)^k-x\varphi^{n-2k+1}}+\frac{1}{-1^{n+1}-x\varphi^{-n-1}}}{\sqrt5^{n+1}}= \frac{\frac{{}_{n+1}C_0}{(-1)^0-x\varphi^{n-2×0+1}}+\sum_{k=1}^n\frac{{}_{n+1}C_k}{(-1)^k-x\varphi^{n-2k+1}}+\frac{{}_{n+1}C_{n+1}}{-1^{n+1}-x\varphi^{n-2(n+1)+1}}}{\sqrt5^{n+1}}= \frac{\sum_{k=0}^{n+1}\frac{{}_{n+1}C_k}{(-1)^k-x\varphi^{n-2k+1}}}{\sqrt5^{n+1}}= \frac{\sum_{k=0}^{n+1}\frac{{}_{n+1}C_k}{(-1)^k-x\varphi^{(n+1)-2k}}}{\sqrt5^{n+1}}= g_{n+1}(x) }$