\documentclass[a4paper]{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{geometry}
\usepackage{tikz}
\usepackage{float}
\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,calc}
\geometry{left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm}
\usepackage{tikz}
\usepackage{pgfplots}
\pgfplotsset{compat=newest, ticks=none}
\usepackage{enumitem}
\newlist{properties}{enumerate}{2}
\setlist[properties]{label=Proprietà \arabic*.,itemindent=*}
\usepackage{tkz-euclide}
\usepackage{booktabs}
\usepackage{amsmath}
\usepackage{multirow}
\usepackage{array}
\newcolumntype{C}{>{\hfil$}p{1.5cm}<{$\hfil}}
\usepackage{mathtools}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\dom}{dom}
\date{}
\title{Notes}
\author{Myself}
\begin{document}
\maketitle
\subsection{Sommatorie}
\begin{equation}
a_{1}+a_{2}+\dots+a_{n}= \sum_{i=1}^{n}a_{i}
\end{equation}
\begin{properties}
\item Prodotto per una costante:
\begin{equation}
\sum_{k=1}^{n}(c \cdot a_{k}) = c \sum_{k=1}^{n} a_{k}
\end{equation}
\item Sommatoria con termine costante:
\begin{equation}
\sum_{k=1}^{n}c = c \cdot n
\end{equation}
\begin{equation}
\sum_{k=m}^{n}(c) = c\cdot (n-m+1)
\end{equation}
\item Sommatoria di sommatorie:
\begin{equation}
\sum_{k=1}^{n}a_{k} \pm \sum_{k=1}^{n}b_{k} = \sum_{k=1}^{n} (a_{k} \pm b_{k})
\end{equation}
\item Scomposizione di una sommatoria:
\begin{equation}
\sum_{k=1}^{n}a_{k} + \sum_{k=n+1}^{n+m}a_{k} = \sum_{k=1}^{n+m} a_{k}
\end{equation}
\item Traslazione di indici
\begin{equation}
\sum_{k=1}^{n}a_{k} = \sum_{k=1+m}^{n+m} a_{k-m}
\end{equation}
\item Riflessione di indici:
\begin{equation}
\sum_{k=1}^{n}a_{k} = \sum_{k=1}^{n} a_{n-k+1} = \sum_{k=0}^{n-1} a_{n-k}
\end{equation}
\item Doppia sommatoria:
\begin{equation}
a_{1,1}+a_{1,2}+\dots+a_{2,1}+a_{2,2}+\dots+a_{n,m}= \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i,j} =\sum_{j=1}^{m} \sum_{i=1}^{n} a_{i,j}
\end{equation}
oppure:
\begin{equation}
\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}b_{j}
\end{equation}
\item Scomposizione doppia sommatoria:
\begin{equation}
\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}b_{j} = \sum_{i=1}^{n} \left( a_{i} \sum_{j=1}^{m} b_{j} \right) = \sum_{i=1}^{n} a_{i} \sum_{j=1}^{m} b_{j}
\end{equation}
\item Cambio d'ordine
\begin{equation}
\sum_{i=1}^{n} \sum_{j=i}^{n} a_{i,j} = \sum_{j=1}^{n} \sum_{i=1}^{j} a_{i,j}
\end{equation}
\item Quadrato di sommatoria
\begin{equation}
\left( \sum_{k=1}^{n}\cdot a_{k} \right) ^{2} = \left( \sum_{k=1}^{n} a_{k} \right) \left( \sum_{k=1}^{n} a_{k} \right) = \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}a_{j} = \sum_{i=1}^{n}a_{i}^{2} + \sum_{i,j=1, i\neq j}^{n}a_{i}a_{j} = \sum_{i=1}^{n}a_{i}^{2} + 2\sum_{i,j=1, i < j}^{n}a_{i}a_{j}
\end{equation}
\end{properties}
\end{document}
我尝试使用\centering，在等式行开头使用\begin{align}和符号，并使用代替，尝试使用，在 内。有什么建议吗？该文档是一篇边距为 2.5 厘米的文章。&\beign{gather}\begin{equation}\begin{aligned}\begin{equation}
答案1
正如 daleif 在其评论中指出的那样，列表中的方程式相对于列表的行居中。由于列表通常有一个左边距，因此与其他方程式相比，这会导致方程式看起来向右移动。
你可以使用 的功能，enumitem在每个数学公式的开头添加负字距。但是，我认为你应该简单地拆分公式，因为它太长了，一行写不完。无论如何，这似乎是你想要的：
\documentclass{article}
\usepackage[a4paper,margin=1in]{geometry}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{enumitem}
\newlist{properties}{enumerate}{2}
\setlist[properties]{label=Proprietà \arabic*.,itemindent=*,before={\everydisplay{\kern-\leftmargin}}}
\usepackage{lipsum}
\begin{document}
\lipsum[66]
\[a=b\]
Here a standard list, showing the difference:
\begin{itemize}
\item blabla\[a=b\]
\item blabla\[a=b\]
\end{itemize}
And here the ``corrected'' version of your list:
\begin{properties}
\item blabla\[a=b\]
\item blabla\[a=b\]
\item Quadrato di sommatoria
\begin{equation}
\begin{aligned}[b]
\biggl( \sum_{k=1}^{n} a_{k} \biggr)^{\!\!2}
&= \biggl( \sum_{k=1}^{n} a_{k} \biggr) \biggl( \sum_{k=1}^{n} a_{k} \biggr)
= \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}a_{j} \\
&= \sum_{i=1}^{n}a_{i}^{2} + \sum_{\substack{i,j=1\\i\neq j}}^{n}a_{i}a_{j}
= \sum_{i=1}^{n}a_{i}^{2} + 2\sum_{\substack{i,j=1\\i < j}}^{n}a_{i}a_{j}
\end{aligned}
\end{equation}
\end{properties}
\end{document}
答案2
如果您希望显示相对于整体边距居中而不是相对于列表中的线宽居中，您可以执行与 AMS 类相同的操作（尽管人们更常见的是要求相反的操作）。
\documentclass[a4paper]{article}
\usepackage{geometry}
\usepackage{amsmath}
\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{calc,trees,positioning,arrows,fit,shapes,calc}
\usepackage{tkz-euclide}
\usepackage{pgfplots}
\usepackage{float}
\usepackage{enumitem}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{array}
\geometry{left=2.5cm,right=2.5cm,top=2.5cm,bottom=2.5cm}
\pgfplotsset{compat=newest, ticks=none}
%\newcolumntype{C}{>{\hfil$}p{1.5cm}<{$\hfil}}
\newcolumntype{C}{>{$}w{c}{1.5cm}<{$}}
\newlist{properties}{enumerate}{2}
\setlist[properties]{label=Proprietà \arabic*.,itemindent=*}
\DeclareMathOperator{\sign}{sign}
\DeclareMathOperator{\im}{im}
\DeclareMathOperator{\dom}{dom}
\providecommand\fullwidthdisplay{%
\setlength{\displayindent}{0pt}%
\setlength{\displaywidth}{\columnwidth}%
}
\everydisplay\expandafter{\expandafter\fullwidthdisplay\the\everydisplay}
\date{}
\title{Notes}
\author{Myself}
\begin{document}
\maketitle
\subsection{Sommatorie}
\begin{equation}
a_{1}+a_{2}+\dots+a_{n}= \sum_{i=1}^{n}a_{i}
\end{equation}
\begin{properties}
\item Prodotto per una costante:
\begin{equation}
\sum_{k=1}^{n}(c \cdot a_{k}) = c \sum_{k=1}^{n} a_{k}
\end{equation}
\item Sommatoria con termine costante:
\begin{equation}
\sum_{k=1}^{n}c = c \cdot n
\end{equation}
\begin{equation}
\sum_{k=m}^{n}(c) = c\cdot (n-m+1)
\end{equation}
\item Sommatoria di sommatorie:
\begin{equation}
\sum_{k=1}^{n}a_{k} \pm \sum_{k=1}^{n}b_{k} = \sum_{k=1}^{n} (a_{k} \pm b_{k})
\end{equation}
\item Scomposizione di una sommatoria:
\begin{equation}
\sum_{k=1}^{n}a_{k} + \sum_{k=n+1}^{n+m}a_{k} = \sum_{k=1}^{n+m} a_{k}
\end{equation}
\item Traslazione di indici
\begin{equation}
\sum_{k=1}^{n}a_{k} = \sum_{k=1+m}^{n+m} a_{k-m}
\end{equation}
\item Riflessione di indici:
\begin{equation}
\sum_{k=1}^{n}a_{k} = \sum_{k=1}^{n} a_{n-k+1} = \sum_{k=0}^{n-1} a_{n-k}
\end{equation}
\item Doppia sommatoria:
\begin{equation}
a_{1,1}+a_{1,2}+\dots+a_{2,1}+a_{2,2}+\dots+a_{n,m}= \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i,j} =\sum_{j=1}^{m} \sum_{i=1}^{n} a_{i,j}
\end{equation}
oppure:
\begin{equation}
\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}b_{j}
\end{equation}
\item Scomposizione doppia sommatoria:
\begin{equation}
\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}b_{j} = \sum_{i=1}^{n} \left( a_{i} \sum_{j=1}^{m} b_{j} \right) = \sum_{i=1}^{n} a_{i} \sum_{j=1}^{m} b_{j}
\end{equation}
\item Cambio d'ordine
\begin{equation}
\sum_{i=1}^{n} \sum_{j=i}^{n} a_{i,j} = \sum_{j=1}^{n} \sum_{i=1}^{j} a_{i,j}
\end{equation}
\item Quadrato di sommatoria
\begin{equation}
\left( \sum_{k=1}^{n}\cdot a_{k} \right) ^{2} = \left( \sum_{k=1}^{n} a_{k} \right) \left( \sum_{k=1}^{n} a_{k} \right) = \sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}a_{j} = \sum_{i=1}^{n}a_{i}^{2} + \sum_{i,j=1, i\neq j}^{n}a_{i}a_{j} = \sum_{i=1}^{n}a_{i}^{2} + 2\sum_{i,j=1, i < j}^{n}a_{i}a_{j}
\end{equation}
\end{properties}
\end{document}
我重新组织了文档前言，将包分成几组，并在包加载后进行设置。请注意C列类型的不同定义。
小心使用\left和\right，因为它们可能会产生过大的围栏（参见“Proprietà 8”）。对于“Proprietà 10”，您可以使用
\item Quadrato di sommatoria
\begin{equation}
\biggl(\, \sum_{k=1}^{n}\cdot a_{k} \biggr) ^{\!2} =
\biggl(\, \sum_{k=1}^{n} a_{k} \biggr) \biggl(\, \sum_{k=1}^{n} a_{k} \biggr) =
\sum_{i=1}^{n} \sum_{j=1}^{m} a_{i}a_{j} =
\sum_{i=1}^{n}a_{i}^{2} + \sum_{\substack{i,j=1 \\ i\neq j}}^{n} a_{i}a_{j} =
\sum_{i=1}^{n}a_{i}^{2} + 2\sum_{\substack{i,j=1 \\ i < j}}^{n}a_{i}a_{j}
\end{equation}
请注意，\,以\biggl(避免发生冲突。
