cases我在环境中多次使用该环境align，输入了相当长的方程式。这会产生很多空白，我想将其消除。
任何关于如何在环境中分页的建议cases或合适的替代方案都将非常有帮助。更具体地说，我知道在序言中输入 \allowdisplaybreaks 不会破坏案例环境（如以下 MWE 所示）。
\documentclass[11pt,a4paper]{amsart}
\allowdisplaybreaks
\usepackage{enumerate,amssymb,amsmath}
\begin{document}
\begin{align*}
&\text{something}\\
&=
\begin{cases}
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if A;}\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if B.}\\
\end{cases}
\\
&=
\begin{cases}
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if A;}\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if B.}\\
\end{cases}
\\
&=
\begin{cases}
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if A;}\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
\\
\displaystyle{+
\sum_{i=1}^{\frac{1}{2}(k-6)}
\frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)}
\binom{\frac{n}{2}}{i}
\binom{\frac{n}{4}}{k-6i-1}v^{k-2i}}
&\text{if B.}\\
\end{cases}
\end{align*}
\end{document}
答案1
在这种情况下，我倾向于重新考虑我的符号，而不是寻找一个TeX基于的解决方案。即使你找到了一种创建可以跨页的案例环境的方法，结果看起来也不会很好，而且可读性也会很差。如果没有看到你的实际方程式，很难提出具体的建议，但如果你展示的术语反复出现，我倾向于定义
r_{nk} = \frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)},
因为这样可以节省很多空间。
答案2
我正在回答这个问题的“合适的替代方案”部分。我也遇到过同样的问题，到目前为止我能找到的最佳答案是以下问题的答案：Tikz - 如何在长桌上覆盖装饰
诚然，这远非理想，但它是一种可能的替代方案。