Update 文章 “使用归纳法证明二项式定理”
All checks were successful
Build and Deploy Qingshuige / build-deploy (push) Successful in 1m30s

This commit is contained in:
2026-04-16 21:14:26 +08:00
parent 760476ceb4
commit d2e41f81e2

View File

@@ -22,14 +22,14 @@ $$
试图使左右两边相等,需要使得左边分母为$k!(n+1-k)!$,所以通分得:
$$
\begin{align}
\begin{aligned}
& \frac{n!}{(k-1)!(n-k+1)!} + \frac{n!}{k!(n-k)!} \\
&= \frac{n!k}{k!(n-k+1)!} + \frac{n!(n-k+1)}{k!(n-k+1)!} \\
&= \frac{n!k+n!(n-k+1)}{k!(n-k+1)!} \\
&= \frac{n!(n+1)}{k!(n-k+1)!} \\
&= \frac{(n+1)!}{k!(n+1-k)!} \\
&= \binom{n+1}{k}. \tag*{$\blacksquare$}
\end{align}
&= \binom{n+1}{k}. \blacksquare
\end{aligned}
$$
随后证明二项式定理
@@ -77,9 +77,9 @@ $$
所以,
$$
\begin{align}
(a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k}b^{k}. \tag*{$\blacksquare$}
\end{align}
\begin{aligned}
(a+b)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} a^{n+1-k}b^{k}. \blacksquare
\end{aligned}
$$
虽然在这里写得洋洋洒洒,但是说实话,在下天资愚钝,第一次见到这个证明,还是没那么容易理解的. 而且很有可能,现在证得出来,将来某天就会忘掉. 所以数学的学习是一个不断积累的过程,现在吃下的每一口,将来都会成长为构成身体的肌肉和骨骼.