フィボナッチ数列の総和
フィボナッチ数列の総和
フィボナッチ数の総和について次が成り立つ。
フィボナッチ数の総和について次が成り立つ。
(1)通常
\[ \sum_{k=0}^{n}F_{k}=F_{n+2}-1 \](2)奇数
\[ \sum_{k=1}^{n}F_{2k-1}=F_{2n} \](3)偶数
\[ \sum_{k=0}^{n}F_{2k}=F_{2n+1}-1 \](4)2乗
\[ \sum_{k=0}^{n}F_{k}^{2}=F_{n}F_{n+1} \](1)
\begin{align*} \sum_{k=0}^{6}F_{k} & =F_{6+2}-1\\ & =F_{8}-1\\ & =20 \end{align*}(2)
\begin{align*} \sum_{k=1}^{3}F_{2k-1} & =F_{2\cdot3}\\ & =F_{6}\\ & =8 \end{align*}(3)
\begin{align*} \sum_{k=0}^{3}F_{2k} & =F_{2\cdot3+1}-1\\ & =F_{7}-1\\ & =12 \end{align*}(4)
\begin{align*} \sum_{k=0}^{3}F_{k}^{2} & =F_{3}F_{3+1}\\ & =F_{3}F_{4}\\ & =2\cdot3\\ & =6 \end{align*}(1)
\begin{align*} \sum_{k=0}^{n}F_{k} & =F_{0}+F_{1}+\sum_{k=2}^{n}F_{k}\\ & =F_{0}+F_{1}+\sum_{k=2}^{n}\left(F_{k-1}+F_{k-2}\right)\\ & =F_{0}+F_{1}+\sum_{k=1}^{n-1}F_{k}+\sum_{k=0}^{n-2}F_{k}\\ & =F_{0}+F_{1}+\LHS-F_{0}-F_{n}+\LHS-F_{n-1}-F_{n}\\ & =F_{1}+2\LHS-F_{n}-F_{n+1}\\ & =F_{1}+2\LHS-F_{n+2}\\ & =F_{n+2}-F_{1}\\ & =F_{n+2}-1 \end{align*}(2)
\begin{align*} \sum_{k=1}^{n}F_{2k-1} & =F_{1}+\sum_{k=2}^{n}F_{2k-1}\\ & =F_{1}+\sum_{k=2}^{n}\left(F_{2k-2}+F_{2k-3}\right)\\ & =F_{1}+\sum_{k=1}^{n-1}\left(F_{2k}+F_{2k-1}\right)\\ & =F_{1}+\sum_{k=1}^{2\left(n-1\right)}F_{k}\\ & =F_{1}+\sum_{k=1}^{2\left(n-1\right)}F_{k}\\ & =F_{1}+F_{2\left(n-1\right)+2}-F_{1}-F_{0}\\ & =F_{2n}-F_{0}\\ & =F_{2n} \end{align*}(3)
\begin{align*} \sum_{k=0}^{n}F_{2k} & =F_{0}+\sum_{k=1}^{n}F_{2k}\\ & =F_{0}+\sum_{k=1}^{n}\left(F_{2k-1}+F_{2k-2}\right)\\ & =F_{0}+\sum_{k=0}^{n-1}\left(F_{2k+1}+F_{2k}\right)\\ & =F_{0}+\sum_{k=0}^{2n-1}F_{k}\\ & =F_{2n+1}-F_{1}+F_{0}\\ & =F_{2n+1}-1 \end{align*}(4)
\begin{align*} \sum_{k=0}^{n}F_{k}^{2} & =F_{0}^{2}+\sum_{k=1}^{n}F_{k}^{2}\\ & =F_{0}^{2}+\sum_{k=1}^{n}F_{k}\left(F_{k+1}-F_{k-1}\right)\\ & =F_{0}^{2}+\sum_{k=1}^{n}\left(F_{k}F_{k+1}-F_{k-1}F_{k}\right)\\ & =F_{n}F_{n+1}-F_{0}F_{1}+F_{0}^{2}\\ & =F_{n}F_{n+1} \end{align*}ページ情報
タイトル | フィボナッチ数列の総和 |
URL | https://www.nomuramath.com/qxlfil4b/ |
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フィボナッチ数列の加法定理
\[
F_{m+n}=F_{m-1}F_{n}+F_{m}F_{n+1}
\]
フィボナッチ数列の組み合せ論的解釈
$n$段の階段を1段または2段ずつ登るときの登り方は$F_{n+1}$通り。
フィボナッチ数列同士の最大公約数
\[
\gcd\left(F_{m},F_{n}\right)=F_{\gcd\left(m,n\right)}
\]
フィボナッチ数列の商の極限
\[
\lim_{n\rightarrow\infty}\frac{F_{n+1}}{F_{n}}=\phi
\]