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)

${\displaystyle \sum _{m=0}^{\infty }(F_m{)}^n{x}^m=\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-x{\phi }^{n-2k}}}{{\sqrt{5}}^n}}$

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

${\displaystyle f_0(x)=\frac{1}{1-x}{f}_{n+1}(x)=\frac{f_n(x\phi )-f_n(-\frac{x}{\phi })}{\sqrt{5}}}$
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:
${\displaystyle \frac{g_n(x\phi )-g_n(-\frac{x}{\phi })}{\sqrt{5}}=\frac{\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-(x\phi ){\phi }^{n-2k}}}{{\sqrt{5}}^n}-\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-(\frac{-x}{\phi }){\phi }^{n-2k}}}{{\sqrt{5}}^n}}{\sqrt{5}}=\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}-\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k+x{\phi }^{n-2k-1}}}{{\sqrt{5}}^{n+1}}=\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}+\sum _{k=0}^n\frac{{}_n{C}_{(k+1)-1}}{(-1{)}^{k+1}-x{\phi }^{n-2(k+1)+1}}}{{\sqrt{5}}^{n+1}}=\frac{\sum _{k=0}^n\frac{{}_n{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}+\sum _{k=1}^{n+1}\frac{{}_n{C}_{k-1}}{(-1{)}^k-x{\phi }^{n-2k+1}}}{{\sqrt{5}}^{n+1}}=\frac{\frac{1}{1-x{\phi }^{n+1}}+\sum _{k=1}^n\frac{{}_n{C}_k+{}_n{C}_{k-1}}{(-1{)}^k-x{\phi }^{n-2k+1}}+\frac{1}{-1^{n+1}-x{\phi }^{-n-1}}}{{\sqrt{5}}^{n+1}}=\frac{\frac{1}{1-x{\phi }^{n+1}}+\sum _{k=1}^n\frac{{}_{n+1}{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}+\frac{1}{-1^{n+1}-x{\phi }^{-n-1}}}{{\sqrt{5}}^{n+1}}=\frac{\frac{{}_{n+1}{C}_0}{(-1{)}^0-x{\phi }^{n-2\times 0+1}}+\sum _{k=1}^n\frac{{}_{n+1}{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}+\frac{{}_{n+1}{C}_{n+1}}{-1^{n+1}-x{\phi }^{n-2(n+1)+1}}}{{\sqrt{5}}^{n+1}}=\frac{\sum _{k=0}^{n+1}\frac{{}_{n+1}{C}_k}{(-1{)}^k-x{\phi }^{n-2k+1}}}{{\sqrt{5}}^{n+1}}=\frac{\sum _{k=0}^{n+1}\frac{{}_{n+1}{C}_k}{(-1{)}^k-x{\phi }^{(n+1)-2k}}}{{\sqrt{5}}^{n+1}}=g_{n+1}(x)}$