From 910eacbab9f6a521981c993e4ff073e630932df2 Mon Sep 17 00:00:00 2001
From: idealist999 <2974730459@qq.com>
Date: Wed, 31 Dec 2025 20:53:23 +0800
Subject: [PATCH] vault backup: 2025-12-31 20:53:23

---
 编写小组/试卷/0103高数模拟试卷.md | 6 +++---
 1 file changed, 3 insertions(+), 3 deletions(-)

diff --git a/编写小组/试卷/0103高数模拟试卷.md b/编写小组/试卷/0103高数模拟试卷.md
index 68aea5a..824a21b 100644
--- a/编写小组/试卷/0103高数模拟试卷.md
+++ b/编写小组/试卷/0103高数模拟试卷.md
@@ -515,9 +515,9 @@ $$|x_{n+1} - x_n| \leq r |x_n - x_{n-1}| \leq r^2 |x_{n-1} - x_{n-2}| \leq \cdot
 于是对任意正整数 $m > n$，有  
 $$
 \begin{aligned}
-|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\
-&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\
-&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\
+|x_m - x_n| &\leq |x_m - x_{m-1}| + |x_{m-1} - x_{m-2}| + \cdots + |x_{n+1} - x_n| \\[1em]
+&\leq (r^{m-2} + r^{m-3} + \cdots + r^{n-1}) |x_2 - x_1| \\[1em]
+&= r^{n-1} \cdot \frac{1 - r^{m-n}}{1 - r} \cdot |x_2 - x_1| \\[1em]
 &\leq \frac{r^{n-1}}{1 - r} |x_2 - x_1|.
 \end{aligned}
 $$