具體數學之二項式係數1

本章講述的是二項式係數，包含了一大堆記不住的公式@<@

1.$\left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\left\{\begin{array}{l}{\frac{r(r-1) \cdots(r-k+1)}{k(k-1) \cdots(1)}=\frac{r^{k}}{k ! } , k \geqslant 0} \\ {0}, k&lt;0\end{array}\right.$
當k=0時，上述結果爲1
r爲上指標，k爲下指標，表示從r個數裏面取k個的排序

2.帕斯卡三角形（楊輝三角形）

$\left( \begin{array}{l}{r} \\ {0}\end{array}\right)=1, \left( \begin{array}{l}{r} \\ {1}\end{array}\right)=r, \left( \begin{array}{l}{r} \\ {2}\end{array}\right)=\frac{r(r-1)}{2}$

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☆ 考慮式子$\left( \begin{array}{c}{-1} \\ {k}\end{array}\right) \stackrel{?}{=} \left( \begin{array}{c}{-1} \\ {-1-k}\end{array}\right)$

$\left( \begin{array}{c}{-1} \\ {k}\end{array}\right)=\frac{(-1)(-2) \cdots(-k)}{k !}=(-1)^{k}$

$\left( \begin{array}{c}{-1} \\ {-1-k}\end{array}\right)=(-1)^{-1-k}$（由上式可以得出），它是1或者-1
因此上述等式總是相等是錯誤的！

3.吸收等式（5.5）

$\left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\frac{r}{k} \left( \begin{array}{l}{r-1} \\ {k-1}\end{array}\right)$，$整數 k \neq 0$

4.相伴恆等式（5.7）

$(r-k) \left( \begin{array}{l}{r} \\ {k}\end{array}\right)=r \left( \begin{array}{c}{r-1} \\ {k}\end{array}\right)$

$\begin{aligned}(r-k) \left( \begin{array}{c}{r} \\ {k}\end{array}\right) &amp;=(r-k) \left( \begin{array}{c}{r} \\ {r-k}\end{array}\right) ，對稱性\\ &amp;=r \left( \begin{array}{c}{r-1} \\ {r-k-1}\end{array}\right)，吸收等式 \\ &amp;=r \left( \begin{array}{c}{r-1} \\ {k}\end{array}\right) ，對稱性\end{aligned}$

5.加法公式（楊輝三角的性質）（5.8）

$\left( \begin{array}{l}{r} \\ {k}\end{array}\right)=\left( \begin{array}{c}{r-1} \\ {k}\end{array}\right)+\left( \begin{array}{l}{r-1} \\ {k-1}\end{array}\right)$，k是整數

利用定義證明：
$\sum_{k \leqslant n} \left( \begin{array}{c}{r+k} \\ {k}\end{array}\right)=\left( \begin{array}{l}{r} \\ {0}\end{array}\right)+\left( \begin{array}{c}{r+1} \\ {1}\end{array}\right)+\cdots+\left( \begin{array}{c}{r+n} \\ {n}\end{array}\right)$$=\left( \begin{array}{c}{r+n+1} \\ {n}\end{array}\right)$，n是整數

6.關於上指標求和

$\sum_{0 \leqslant k \leqslant n} \left( \begin{array}{l}{k} \\ {m}\end{array}\right)=\left( \begin{array}{l}{0} \\ {m}\end{array}\right)+\left( \begin{array}{l}{1} \\ {m}\end{array}\right)+\cdots+\left( \begin{array}{l}{n} \\ {m}\end{array}\right)$$=\left( \begin{array}{l}{n+1} \\ {m+1}\end{array}\right)$，整數 $m, n \geqslant 0$

$\begin{aligned} \sum_{k \leq n} \left( \begin{array}{c}{m+k} \\ {k}\end{array}\right) &amp;=\sum_{-m \leq k \leqslant n} \left( \begin{array}{c}{m+k} \\ {k}\end{array}\right) \\ &amp;=\sum_{-m \leq k \leqslant n} \left( \begin{array}{c}{m+k} \\ {m}\end{array}\right) \\ &amp;=\sum_{0 \leqslant k \leqslant m+n} \left( \begin{array}{c}{k} \\ {m}\end{array}\right) \\ &amp;=\left( \begin{array}{c}{m+n+1} \\ {m+1}\end{array}\right)=\left( \begin{array}{c}{m+n+1} \\ {n}\end{array}\right) \end{aligned}$

7.上指標反轉（5.14）

$\left( \begin{array}{l}{r} \\ {k}\end{array}\right)=(-1)^{k} \left( \begin{array}{c}{k-r-1} \\ {k}\end{array}\right)$，k是整數

8. $(-1)^{m} \left( \begin{array}{c}{-n-1} \\ {m}\end{array}\right)=(-1)^{n} \left( \begin{array}{c}{-m-1} \\ {n}\end{array}\right)$$=\left( \begin{array}{c}{m+n} \\ {n}\end{array}\right)$，整數 $m, n \geqslant 0$

利用上指標公式，也可以推導出下列式子（帕斯卡三角形一行的部分交替求和）：
$\sum_{k≤m} \left( \begin{array}{l}{r} \\ {k}\end{array}\right)(-1)^{k}=\left( \begin{array}{l}{r} \\ {0}\end{array}\right)-\left( \begin{array}{l}{r} \\ {1}\end{array}\right)+\cdots+(-1)^{m} \left( \begin{array}{l}{r} \\ {m}\end{array}\right)$$=(-1)^{m} \left( \begin{array}{c}{r-1} \\ {m}\end{array}\right), \quad m是整數$

$\begin{aligned} \sum_{k \leqslant m} \left( \begin{array}{c}{r} \\ {k}\end{array}\right)(-1)^{k}=\sum_{k \leqslant m} \left( \begin{array}{c}{k-r-1} \\ {k}\end{array}\right) \\ =\left( \begin{array}{c}{-r+m} \\ {m}\end{array}\right) \\=(-1)^{m} \left( \begin{array}{c}{r-1} \\ {m}\end{array}\right) \end{aligned}$