ARC144D 的一步组合式推导

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})

$\sum_{i = x} ^{k + 1}{(k + 1 - i) * \binom{i - 1}{x - 1}} = \binom{k + 1}{x + 1} \\ \Leftrightarrow \sum_{i = x}^{k}{(k - i)\binom{i - 1}{x - 1}} = \binom{k}{x + 1}$

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})

$\sum_{i = x}^{k}{(k - i) * \binom{i - 1}{x - 1} } \\ = \sum_{i = x} ^ {k}{k \binom{i - 1}{x - 1} - x \binom{i}{x}} \\ = k \sum_{i = x} ^ {k} {\binom{i - 1}{x - 1}} - x \sum_{i = 0}^{k}{\binom{i}{x}} \\ = k * \sum_{i = x - 1} ^{k - 1}{\binom{i}{x - 1}} - x * \sum_{i = 0} ^{k}{\binom{i}{x}} \\ = k * \binom{k}{x} - x * \binom{k + 1}{x + 1} \\ =(k + 1)\binom{k}{x} - x * \binom{k + 1}{x + 1} - \binom{k}{x} \\ =(x + 1)\binom{k + 1}{x+ 1} - x * \binom{k + 1}{x + 1} - \binom{k}{x} \\ = \binom{k + 1}{x+ 1} - \binom{k}{x} \\ = \binom{k+ 1}{x}$

考虑拓展组合意义的证明：