二項式系數
二項式系數
說明:一般情況下\(0^0 = 1\)
對所有的非負整數\(k\)和\(n\)定義的二項式系數\(\binom{n}{k}\),
如果\(k > n\),則\(\binom{n}{k} = 0\),
對所有的\(n\),\(\binom{n}{0} = 1\),
如果\(n\)是是一個正整數,且\(1 \le k \le b\),則\(\binom{n}{k} = \frac{n!}{k! (n - k)!} = \frac{n (n - 1) \cdots (n - k + 1)}{k (k - 1) \cdots 1}\)
有以下一些性質成立
\(\binom{n}{k} = \binom{n}{n - k} \tag{1}\)
顯然
Pascal公式
\(\binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1} \tag{2}\)
根據定義或組合意義可證
\(\binom{n}{0} + \binom{n}{1} + \cdots + \binom{n}{n} = 2^n \ \ \ \ \ \ (n \ge 0) \tag{3}\)
可用有關二進制的組合意義理解
二項式定理
若\(n\)是正整數,對於所有的\(x, y\)有\((x + y)^n = \sum\limits_{k = 0}^n \binom{n}{k} x^{n - k} y^k \tag{4}\)
數學歸納即可
特殊情況
當\(x = y = 1\),即可推出\((3)\)
當\(y = 1\),有\((x + 1)^n = \sum\limits_{k = 0}^n \binom{n}{k} x^k = \sum\limits_{k = 0}^n \binom{n}{n - k} x^k \tag{5}\)
當\(x = 1, y = -1\),推出
\(\sum\limits_{k = 0}^n (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \cdots + (-1)^n \binom{n}{n} = 0 \ \ \ \ \ (n \ge 1) \tag{6}\)
結合\((3)\)即可得\(\binom{n}{0} + \binom{n}{2} + \cdots = \binom{n}{1} + \binom{n}{3} + \cdots = 2^{n - 1} \ \ \ \ \ (n \ge 1) \tag{7}\)
\(k \binom{n}{k} = n \binom{n - 1}{k - 1} \ \ \ \ \ (n,k都是正整數) \tag{8}\)
根據定義可證
\(\sum\limits_{k = 1}^n k\binom{n}{k} = n 2^{n - 1} \ \ \ \ \ (n \ge 1) \tag{9}\)
證明一:
令\(s = 0 \binom{n}{0} + \binom{n}{1} + 2 \binom{n}{2} + \cdots + n \binom{n}{n}\),
將所有項反轉,得\(s = n \binom{n}{n} + \cdots + 2 \binom{n}{2} + \binom{n}{1} + 0 \binom{n}{0}\),
由\((1)\),兩式相加得\(2s = n \sum\limits_{k = 0}^n \binom{n}{k} = n 2^n\),所以\(s = n 2^{n - 1}\)
證明二:
根據\((8)\)和\((3)\)可得\(\sum\limits_{k = 0}^n k\binom{n}{k} = \sum\limits_{k = 1}^n n\binom{n - 1}{k - 1} = n \sum\limits_{k = 0}^{n - 1} \binom{n - 1}{k} = n 2^{n - 1}\)
證明三:
對\((x + 1)^n = \sum\limits_{k = 0}^n \binom{n}{k} x^k \ \ \ \ \ (5)\)兩邊關於\(x\)求導,可得\(n (x + 1)^{n - 1} = \sum\limits_{k = 1}^n k \binom{n}{k} x^{k - 1}\)
代入\(x = 1\)即可
\(\sum\limits_{k = 1}^n k^2 \binom{n}{k} = n (n + 1) 2^{n - 2} \tag{10}\)
二項式系數