组合恒等式整理

1. $\binom{n}{m} = \binom{n}{n-m}$
2. $\binom{n}{m} + \binom{n}{m+1} = \binom{n+1}{m+1}$
3. $\sum_{m=r}^n \binom{m}{r} = \binom{n+1}{r+1}$
4. $\sum_{m=0}^n \binom{n}{m} = 2^n$
5. $\binom{n}{m} \binom{m}{k} = \binom{n}{k} \binom{n-k}{m-k}$
   1. $m \binom{n}{m} = n \binom{n-1}{m-1}$
6. $\sum_{r=0}^k \binom{n}{r} \binom{m}{k-r} = \binom{n+m}{k}$
   1. $\sum_{i=0}^n \binom{n}{i}^2 = \sum_{i=0}^n \binom{n}{i} \binom{n}{n-i} = \binom{2n}{n}$

封闭形式：

1. $\frac{1}{(1-x)^{n+1}} = \sum_{k=0}^\infty \binom{n+k}{n} x^k$

   1. $\frac{1}{(1-x)^{n+1}} = (1-x)^{-(n+1)} = \sum_{k=0}^\infty \binom{-n-1}{k} (-x)^k = \sum_{k=0}^\infty (-1)^k \binom{k+n}{k} (-x)^k = \sum_{k=0}^\infty \binom{n+k}{n} x^k$

2. $\frac{1}{\sqrt{1-4x}} = \sum_{k=0}^\infty \binom{2k}{k} x^k$

   1. $$(1-x)^{-\frac{1}{2}} = \sum_{k=0}^\infty \binom{-\frac{1}{2}}{k} (-x)^k\\ = \sum_{k=0}^\infty \frac{(-1)(-3) \cdots (-(2k-1))}{2^k k!} (-x)^k\\ = \sum_{k=0}^\infty (-1)^k \frac{\frac{(2k)!}{2^k k!}}{2^k k!} (-x)^k\\ = \sum_{k=0}^\infty \binom{2k}{k} 4^{-k} x^k\\ \frac{1}{\sqrt{1-4x}} = (1-4x)^{-\frac{1}{2}} = \sum_{k=0}^\infty \binom{2k}{k} 4^{-k} (4x)^k\\ = \sum_{k=0}^\infty \binom{2k}{k} x^k$$

3. $\frac{1 - \sqrt{1-4x}}{2x} = \sum_{k\ge 0} \frac{1}{k+1} \binom{2k}{k} x^k$

   1. $$(1-x)^{\frac{1}{2}} = \sum_{k=0}^\infty \binom{\frac{1}{2}}{k} (-x)^k\\ = \sum_{k=0}^\infty \frac{1(-1)(-3)(-5)\cdots (-(2k-3))}{2^k k!}(-x)^k\\ = - \sum_{k=0}^\infty \frac{\frac{(2k-2)!}{2^{(k-1)}(k-1)!}}{2^kk!}x^k\\ = - 2 \sum_{k=0}^\infty \frac{1}{k} \cdot\binom{2k-2}{k-1} \cdot 4^{-k} x^k\\ 1 - \sqrt{1-4x} = 2 \sum_{k=1}^\infty \frac{1}{k} \cdot\binom{2k-2}{k-1} \cdot 4^{-k} (4x)^k\\ = 2 \sum_{k=1}^\infty \frac{1}{k} \cdot\binom{2k-2}{k-1} \cdot x^k$$