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如何展开等号两边的求和符号

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如何展开等号两边的求和符号

我想您明白我想做什么，但我不知道如何将等号完全对齐，同时仍保留垂直省略号。

Assume $\displaystyle \sum_{i=1}^{n}\alpha _{i}^{i}=\sum_{i=1}^{n}\beta _{i}^{i}$ for $j=1,\ldots,k-1$.  By definition,
\begin{align} 
\alpha_{1}+\alpha_{2}+\ldots+\alpha_{n} &= \beta_{1}+\beta_{2}+\ldots+\beta_{n}\\ 
\alpha_{1}^2+\alpha_{2}^2+\ldots+\alpha_{n}^2 &= \beta_{1}^2+\beta_{2}^2+\ldots+\beta_{n}^2\\
\vdots+\vdots+\ldots+\vdots &= \vdots+\vdots+\ldots+\vdots\\
\alpha_{1}^{k-1}+\alpha_{2}^{k-1}+\ldots+\alpha_{n}^{k-1}&= \beta_{1}^{k-1}+\beta_{2}^{k-1}+\ldots+\beta_{n}^{k-1}
\end{align}

答案1

我建议使用alignat在等号和加号处对齐的环境。您也可以使用\或\quad等来\vdots对齐它们。

举个例子，也许用&得比必要的还多：

\begin{alignat}{6}
\alpha_{1}&+\alpha_{2}&&+\ldots&&+\alpha_{n} &&= \beta_{1}&&+\beta_{2}&&+\ldots+\beta_{n}\\ 
\alpha_{1}^2&+\alpha_{2}^2&&+\ldots&&+\alpha_{n}^2 &&= \beta_{1}^2&&+\beta_{2}^2&&+\ldots+\beta_{n}^2\\
\vdots\ &+\ \vdots&&+\ldots&&+\ \vdots &&= \ \vdots&&+\ \vdots&&+\ldots+\ \vdots
\end{alignat}

方程式

答案2

\documentclass[a4paper]{article}
\usepackage{amsmath,calc}

\newcommand{\xalpha}[3][\alpha]{%
  \makebox[\widthof{$\alpha^{k-1}$}]{$#1_{#2}^{#3}$}}
\newcommand{\xbeta}[3][\beta]{%
  \makebox[\widthof{$\beta^{k-1}$}]{$#1_{#2}^{#3}$}}
\newcommand{\lvdots}{\smash{\raisebox{-.5ex}{$\vdots$}}}

\begin{document}

\begin{gather}
\xalpha{1}{}+\xalpha{2}{}+\ldots+\xalpha{n}{} =
  \xbeta{1}{}+\xbeta{2}{}+\ldots+\xbeta{n}{}\\
\xalpha{1}{2}+\xalpha{2}{2}+\ldots+\xalpha{n}{2} =
  \xbeta{1}{2}+\xbeta{2}{2}+\ldots+\xbeta{n}{2}\\
\xalpha[\lvdots]{}{}+\xalpha[\lvdots]{}{}+\ldots+\xalpha[\lvdots]{}{} =
  \xbeta[\lvdots]{}{}+\xbeta[\lvdots]{}{}+\ldots+\xbeta[\lvdots]{}{}\\
\xalpha{1}{k-1}+\xalpha{2}{k-1}+\ldots+\xalpha{n}{k-1} =
  \xbeta{1}{k-1}+\xbeta{2}{k-1}+\ldots+\xbeta{n}{k-1}
\end{gather}
\end{document}

\xalpha我通过和将所有加数设置为相同宽度，将实际加数置于与或\xbeta一样宽的框中。需要稍微降低一点。\alpha^{k-1}\beta^{k-1}\vdots

人们可能会使用更复杂的宏，考虑到所有半线都是同一类型。

在此处输入图片描述

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