在对齐环境中与对齐环境对齐

Question

我有一个多项式展开式中二项式系数和的表达式(x - y)^{4}。（1代替x和y。）这个和的等价表达式要写成几行。有术语

+ \left[(-1)^{3} \binom{3}{2} + (-1)^{3} \binom{3}{3}\right] + (-1)^{4}\binom{3}{3}

在第二行。我希望将第二行中的表达式分成两行，并且我+希望

+ \left[(-1)^{1} \binom{3}{0} + (-1)^{1} \binom{3}{1}\right]

与上一个表达式中的垂直对齐+。（我希望在下一行中有类似的分隔。）

\documentclass{amsart}
\usepackage{amsmath}



\setlength{\oddsidemargin}{0.0in}
\setlength{\evensidemargin}{0.0in} \setlength{\textwidth}{6.1in}
\setlength{\topmargin}{0.0in} \setlength{\textheight}{9in}



\begin{document}


\begin{align*}
\sum_{i=0}^{4} (-1)^{i} \binom{4}{i}
&= (-1)^{0} \binom{4}{0} + (-1)^{1} \binom{4}{1} + (-1)^{2} \binom{4}{2} + (-1)^{3} \binom{4}{3} + (-1)^{4} \binom{4}{4} \\
&\begin{alignedat}{2}
=(-1)^{0} \binom{3}{0} + \left[(-1)^{1} \binom{3}{0} + (-1)^{1} \binom{3}{1}\right]
+ \left[(-1)^{2} \binom{3}{1} + (-1)^{2} \binom{3}{2}\right] \\
&+ \left[(-1)^{3} \binom{3}{2} + (-1)^{3} \binom{3}{3}\right] + (-1)^{4}\binom{3}{3} \\
= \left[(-1)^{0} \binom{3}{0} + (-1)^{1} \binom{3}{0}\right] + \left[(-1)^{1} \binom{3}{1} + (-1)^{2}\binom{3}{1}\right]
+ \left[(-1)^{2} \binom{3}{2} + (-1)^{3}\binom{3}{2}\right] + \left[(-1)^{3} \binom{3}{3} + (-1)^{4}\binom{3}{3}\right] \\
&= 0. \ \rule{1.5ex}{1.5ex}
\end{alignedat}
\end{align*}

\end{document}

Answer 1

我建议您在环境中使用几个aligned环境align*。

\documentclass{amsart}
\usepackage{amssymb} % for "\blacksquare" macro

\setlength{\oddsidemargin}{0.0in}
\setlength{\evensidemargin}{0.0in} 
\setlength{\textwidth}{6.1in}
\setlength{\topmargin}{0.0in} 
\setlength{\textheight}{9in}

\begin{document}

\begin{align*}
\sum_{i=0}^{4} (-1)^{i} \binom{4}{i}
&= (-1)^{0} \binom{4}{0} + (-1)^{1} \binom{4}{1} + (-1)^{2} \binom{4}{2} + (-1)^{3} \binom{4}{3} + (-1)^{4} \binom{4}{4} \\
&=(-1)^{0} \binom{3}{0} 
\!\begin{aligned}[t] 
&+ \left[(-1)^{1} \binom{3}{0} + (-1)^{1} \binom{3}{1}\right]
+ \left[(-1)^{2} \binom{3}{1} + (-1)^{2} \binom{3}{2}\right] \\
&+ \left[(-1)^{3} \binom{3}{2} + (-1)^{3} \binom{3}{3}\right] + (-1)^{4}\binom{3}{3} 
\end{aligned}\\
&= \!\begin{aligned}[t]
  &\left[(-1)^{0} \binom{3}{0} + (-1)^{1} \binom{3}{0}\right] + \left[(-1)^{1} \binom{3}{1} + (-1)^{2}\binom{3}{1}\right]\\
+ &\left[(-1)^{2} \binom{3}{2} + (-1)^{3}\binom{3}{2}\right] + \left[(-1)^{3} \binom{3}{3} + (-1)^{4}\binom{3}{3}\right]
\end{aligned} \\
&= 0. \quad\blacksquare
\end{align*}

\end{document}

---

附录以解决 OP 的后续请求。修订后的解决方案使用单个align*环境和几个不同的\phantom指令；一个指令使用了四次，另一个指令使用了两次。

\documentclass{amsart}
\usepackage{amssymb} % for "\blacksquare" macro

\setlength{\oddsidemargin}{0.0in}
\setlength{\evensidemargin}{0.0in} 
\setlength{\textwidth}{6.1in}
\setlength{\topmargin}{0.0in} 
\setlength{\textheight}{9in}

%% Two "phantoms"
\newcommand\phA{\phantom{\bigg[}}
\newcommand\phB{\phantom{{}=(-1)^{0} \binom{3}{0}}}
\begin{document}

\begin{align*}
\sum_{i=0}^{4} (-1)^{i} \binom{4}{i}
&=\phA 
   (-1)^{0} \binom{4}{0} + (-1)^{1} \binom{4}{1} 
 + (-1)^{2} \binom{4}{2} + (-1)^{3} \binom{4}{3} + (-1)^{4} \binom{4}{4}\\
&=\phA 
   (-1)^{0} \binom{3}{0} 
 + \biggl[(-1)^{1} \binom{3}{0} + (-1)^{1}\binom{3}{1}\biggr]
 + \biggl[(-1)^{2} \binom{3}{1} + (-1)^{2}\binom{3}{2}\biggr]\\
&\phA\phB
 + \biggl[(-1)^{3} \binom{3}{2} + (-1)^{3}\binom{3}{3}\biggr] 
 + (-1)^{4}\binom{3}{3} \\
&= \biggl[(-1)^{0} \binom{3}{0} + (-1)^{1}\binom{3}{0}\biggr] 
 + \biggl[(-1)^{1} \binom{3}{1} + (-1)^{2}\binom{3}{1}\biggr]\\
&\phA\phB
 + \biggl[(-1)^{2} \binom{3}{2} + (-1)^{3}\binom{3}{2}\biggr] 
 + \biggl[(-1)^{3} \binom{3}{3} + (-1)^{4}\binom{3}{3}\biggr]\\
&= 0. \quad\blacksquare
\end{align*}

\end{document}