位置 1 和位置 2 在同一列,我试过
\begin{align}
x_n&=\begin{aligned}[t]
&\left(1 + \frac{1}{n}\right)^n = \binom{n}{0}\left(\frac{1}{n}\right)^01^n +
\binom{n}{1}\left(\frac{1}{n}\right)^11^{n-1}+\binom{n}{2}\left(\frac{1}
{n}\right)^21^{n-2}+\\
&+\binom{n}{3}\left(\frac{1}{n}\right)^31^{n-3}+\cdots+\binom{n}{k}\left(\frac{1}
{n}\right)^k1^{n-k}+\cdots+\binom{n}{n}\left(\frac{1}{n}\right)^n1^0
\end{aligned}\nonumber\\
&=\begin{aligned}[t]
&1 + 1 + \frac{1}{2!}\left(1 - \frac{1}{n}\right) + \frac{1}{3!}\left(1 - \frac{1}
{n}\right)\left(1 - \frac{2}{n}\right) + \cdots\\
&+ \frac{1}{k!}\left(1 - \frac{1}{n}\right)\cdots\left(1 - \frac{k-1}
{n}\right)+\cdots+\frac{1}{n!}\left(1 - \frac{1}{n}\right)\cdots\left(1-\frac{n-1}
{n}\right).
\end{aligned}
\end{align}
答案1
我建议你在环境内使用单个aligned[b]
环境equation
。这样,只会生成一个方程编号。[b]
定位说明符确保该方程编号将放置在底行。
\documentclass{article} % or some other suitable document class
\usepackage{amsmath} % for 'aligned' environment
\usepackage{mleftright} % see https://www.ctan.org/pkg/mleftright
\mleftright
\begin{document}
\begin{equation}
\begin{aligned}[b]
x_n
&=\left(1 + \frac{1}{n}\right)^n \\
&= \binom{n}{0} \left(\frac{1}{n}\right)^{\!0} 1^n +
\binom{n}{1} \left(\frac{1}{n}\right)^{\!1} 1^{n-1}+
\binom{n}{2} \left(\frac{1}{n}\right)^{\!2} 1^{n-2} \\
&\quad+
\binom{n}{3} \left(\frac{1}{n}\right)^{\!3} 1^{n-3}+\dots+
\binom{n}{k} \left(\frac{1}{n}\right)^{\!k} 1^{n-k}+\dots+
\binom{n}{n} \left(\frac{1}{n}\right)^{\!n} 1^0 \\
&=1 + 1 +
\frac{1}{2!} \left(1 - \frac{1}{n}\right) +
\frac{1}{3!} \left(1 - \frac{1}{n}\right)
\left(1 - \frac{2}{n}\right) + \dotsb\\
&\quad+
\frac{1}{k!} \left(1 - \frac{1}{n} \right) \dotsb
\left(1 - \frac{k-1}{n}\right)+\dots+
\frac{1}{n!} \left(1 - \frac{1}{n} \right) \dotsb
\left(1 - \frac{n-1}{n}\right) .
\end{aligned}
\end{equation}
\end{document}