我们将证明该恒等式
\[\sum_{i = x} ^{k + 1}{(k + 1 - i) * \binom{i - 1}{x - 1}} = \binom{k + 1}{x + 1} \\ \Leftrightarrow \sum_{i = x}^{k}{(k - i)\binom{i - 1}{x - 1}} = \binom{k}{x + 1} \]
可以运用组合意义：\(k\) 个物品选择 \(x + 1\)个物品，枚举倒数第二个位置在哪里
\(\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box\)
代数推导：
\[\sum_{i = x}^{k}{(k - i) * \binom{i - 1}{x - 1} } \\ = \sum_{i = x} ^ {k}{k \binom{i - 1}{x - 1} - x \binom{i}{x}} \\ = k \sum_{i = x} ^ {k} {\binom{i - 1}{x - 1}} - x \sum_{i = 0}^{k}{\binom{i}{x}} \\ = k * \sum_{i = x - 1} ^{k - 1}{\binom{i}{x - 1}} - x * \sum_{i = 0} ^{k}{\binom{i}{x}} \\ = k * \binom{k}{x} - x * \binom{k + 1}{x + 1} \\ =(k + 1)\binom{k}{x} - x * \binom{k + 1}{x + 1} - \binom{k}{x} \\ =(x + 1)\binom{k + 1}{x+ 1} - x * \binom{k + 1}{x + 1} - \binom{k}{x} \\ = \binom{k + 1}{x+ 1} - \binom{k}{x} \\ = \binom{k+ 1}{x} \]
考虑拓展组合意义的证明：
\[\sum_{i = 0}^{n}{\binom{i}{a}\binom{n - i} {b}} = \binom{n + 1}{a + b + 1} \]
可以考虑第\(a+1\)个人的位置
