两个组合数不等式

青青子衿

\[
\begin{array}{cc}
\displaystyle\binom{2n}{n}\!\!\!&\ge&\displaystyle2^n\left(1+\frac{1}{\sqrt[n-1]{n}}\right)^{n-1} \\
\\
\displaystyle\binom{2n+1}{n}\!\!\!&\ge&\displaystyle2^n\left(1+\frac{1}{\sqrt[n]{n+1}}\right)^{n}
\end{array}
\quad\quad n\in\mathbb{N_+}
\]

kuing

你的代码：

1. \[
2. \begin{array}{cc}
3. \displaystyle\binom{2n}{n}\!\!\!&\ge&\displaystyle2^n\left(1+\frac{1}{\sqrt[n-1]{n}}\right)^{n-1} \\
4. \\
5. \displaystyle\binom{2n+1}{n}\!\!\!&\ge&\displaystyle2^n\left(1+\frac{1}{\sqrt[n]{n+1}}\right)^{n}
6. \end{array}
7. \quad\quad n\in\mathbb{N_+}
8. \]

Copy the Code

可以简化成：

1. \[
2. \begin{aligned}
3. \binom{2n}{n} & \ge 2^n\left(1+\frac{1}{\sqrt[n-1]{n}}\right)^{n-1} \\
4. \binom{2n+1}{n} & \ge 2^n\left(1+\frac{1}{\sqrt[n]{n+1}}\right)^{n}
5. \end{aligned}
6. \qquad n\in\mathbb N_+
7. \]

Copy the Code

效果：
\[
\begin{aligned}
\binom{2n}{n} & \ge 2^n\left(1+\frac{1}{\sqrt[n-1]{n}}\right)^{n-1} \\
\binom{2n+1}{n} & \ge 2^n\left(1+\frac{1}{\sqrt[n]{n+1}}\right)^{n}
\end{aligned}
\qquad n\in\mathbb N_+
\]