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第九周。

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2022-11-08 21:18:13 +08:00
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04FibonacciNumbers.tex
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@@ -40,7 +40,7 @@ Fibonacci是递推关系的一个典型问题，这个数列本身也有很多
\end{aligned}
\end{equation*}
累加所有的式子，得到
\[F_1 + F_2 + \cdots + F_n = F_{n+2} - F_2 = F_{n+2} -1 \eqper\]
\[F_1 + F_2 + \cdots + F_n = F_{n+2} - F_2 = F_{n+2} -1 \eqper\qedhere\]
\end{proof}
\item 对等式$F_1 + F_3 + \cdots + F_{2n-1} = F_{2n}$：
\begin{proof}
@@ -53,7 +53,7 @@ Fibonacci是递推关系的一个典型问题，这个数列本身也有很多
\end{aligned}
\end{equation*}
累加所有的式子，得到
\[F_1 + F_3 + \cdots + F_{2n-1} = F_{2n} - F_0 = F_{2n} \eqper\]
\[F_1 + F_3 + \cdots + F_{2n-1} = F_{2n} - F_0 = F_{2n} \eqper\qedhere\]
\end{proof}
\item 对等式$F_0 - F_1 + F_2 - F_3 + \cdots - F_{2n-1} + F_{2n} = F_{2n-1} - 1$：
\begin{proof}
@@ -204,7 +204,7 @@ Fibonacci是递推关系的一个典型问题，这个数列本身也有很多
\right.
\end{equation*}
于是有
\[F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right) \eqper\]
\[F_n = \frac{1}{\sqrt{5}}\left(\left(\frac{1 + \sqrt{5}}{2}\right)^n - \left(\frac{1 - \sqrt{5}}{2}\right)^n\right) \eqper \qedhere\]
\end{proof}
\section{Fibonacci数列的性质}