计数 dp

二项式定理

\[(a + b)^n = \sum_{i = 0}^{n} a^ib^{n-i}\binom{n}{i} \]

排列组合基本运算

这里就不再证明，这都是后面推式子变化的根本。

\[\binom{n}{k} = \binom{n}{n - k} \tag{1} \]

\[\binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1} \tag{2} \]

\[\binom{n}{k} \times k = \binom{n - 1}{k - 1} \times n \tag{3} \]

来两道基本推式子变式

• 变下项求和

\[\sum_{i = 0}^{n} \binom{n}{i} = 2^n \tag{1} \]

\[\sum_{i = 0}^{n} i\binom{n}{i}=2^{n-1}n \tag{2} \]

\[\sum_{i = 0}^{n}i^2 \binom{n}{i} = 2^{n - 2}n(n + 1) \tag{3} \]

证明：

\[\begin{gather*} \because (1+1)^n=\sum_{i=0}^{n} \binom{n}{i} 1^i 1^{n-i} = 2^n \\ \therefore \sum_{i = 0}^{n} \binom{n}{i} = 2^n \\ \end{gather*} \]

\[ \begin{gather*} \because i \binom{n}{i} = n \binom{n-1}{i-1} \\ \therefore \sum_{i=0}^{n} i \binom{n}{i} = \sum_{i = 0}^{n} n \binom{n - 1}{i - 1} = n \sum_{i = 0}^{n} \binom{n-1}{i-1} \\ \therefore \sum_{i=0}^{n} i \binom{n}{i} = 2^{n-1}n \end{gather*} \]

\[\begin{gather*} \because i^2 \binom{n}{i} = i \times n \binom{n-1}{i-1}=(i-1+1) \times n \binom{n-1}{i-1} \\ = (i-1)\times n \binom{n-1}{i-1} + n \binom{n-1}{i-1} = n(n-1)\binom{n-2}{i-2} + n \binom{n-1}{i-1} \\ \therefore \sum_{i=0}^n n(n-1)\binom{n-2}{i-2} + n \binom{n-1}{i-1} \\ = 2^{n-2}n(n-1) + 2^{n-1}n = 2^{n-2}n (n-1+2)=2^{n-2}n(n+1) \end{gather*} \]

• 变上项求和

\[\sum_{i = 0}^{n} \binom{i}{k} = \binom{n+1}{k+1} \]

证明: 直接按照 $$\binom{n}{k} = \binom{n - 1}{k} + \binom{n - 1}{k - 1}$$ 一直展开就行了