cases我在環境中多次使用該環境align，輸入相當長的方程式。這會產生很多空白，我想將其刪除。
任何有關如何在環境中分頁cases或合適的替代方案的建議將是最有幫助的。更具體地說，我知道在序言中輸入 \allowdisplaybreaks 不會破壞 case 環境（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
在這種情況下，我會傾向於重新考慮我的符號，而不是尋找一個TeX有基礎的解決方案。即使你找到了一種方法來創建類似案例的環境，可以跨頁，結果看起來也不會很好，而且可讀性也會很差。在沒有看到你的實際方程式的情況下很難提出具體的建議，但是如果你顯示的術語重複出現，我會傾向於定義
r_{nk} = \frac{n^2-2n(k-3i+6)-4i}{n(2k+7i)},
因為這樣可以節省很多空間。
答案2
我正在回答這個問題的“合適的替代方案”部分。我也有同樣的問題，到目前為止我能找到的最佳答案是以下問題的答案：Tikz - 如何在 longtable 上疊加裝飾
誠然，這遠非理想，但它是一種可能的替代方案。