From 393b740fa0e694538356521cbc4d238058d2be19 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=8E=8B=E8=BD=B2=E6=A5=A0?= Date: Tue, 13 Jan 2026 15:42:38 +0800 Subject: [PATCH] vault backup: 2026-01-13 15:42:38 --- .../试卷/线代期末复习模拟/1.14线代测试答案.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md index 6afd7b5..ac44c30 100644 --- a/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md +++ b/编写小组/试卷/线代期末复习模拟/1.14线代测试答案.md @@ -212,7 +212,7 @@ $$F_k=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^k-\left(\frac{1- 对比初始条件：$D_1=1=F_2，D_2=2=F_3$，故 $D_n=F_{n+1}$（斐波那契数列的第 $n+1$ 项）。 -将$ F_{n+1} 代入比内公式$，得： +将$F_{n+1} 代入比内公式$，得： $$D_n=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right]$$