关于广义加法公式的一种证明

对任意事件\(A_i(i=1 \cdots n), (n \geq 2)\),有：

\[P(\bigcup\limits_{i=1}^{n}A_i) = \sum\limits_{i=1}^{n}P(A_i) - \sum\limits_{1 \leq i < j \leq n}^{n}P(A_iA_j) + \cdots +(-1)^{n+1}P(\bigcap\limits_{i=1}^{n}A_i)\]

使用数学归纳法证明如下：
当\(n=2\)时，\(P(A_1 \cup A_2) = P(A_1 \cup (A_2 - A_1A_2)) = P(A_1) + P(A_2 - A_1A_2)[有限可加性] = P(A_1) + P(A_2) - P(A_1A_2)[由于A_1A_2 \subset A_2]\)
当\(n=k\)时，假设:\(P(\bigcup\limits_{i=1}^{k}A_i) = \sum\limits_{i=1}^{k}P(A_i) - \sum\limits_{1 \leq i < j \leq k}^{k}P(A_iA_j) + \cdots +(-1)^{k+1}P(\bigcap\limits_{i=1}^{k}A_i)\)
则当n=k+1时，\(P(\bigcup\limits_{i=1}^{k+1}A_i) = P(A_{k+1}) + P(\bigcup\limits_{i=1}^{k}A_i) - P(\bigcup\limits_{i=1}^{k}A_{k+1}A_i) = P(A_{k+1}) +\sum\limits_{i=1}^{k}P(A_i) - \sum\limits_{1 \leq i < j \leq k}^{k}P(A_iA_j) + \cdots +(-1)^{k+1}P(\bigcap\limits_{i=1}^{k}A_i) - \sum\limits_{i=1}^{k}P(A_{k+1}A_i) + \sum\limits_{1 \leq i < j \leq k}^{k}P(A_{k+1}A_iA_j) - \cdots -(-1)^{k+1}P(A_{k+1}\bigcap\limits_{i=1}^{k}A_i) = \sum\limits_{i=1}^{k+1}P(A_i) - \sum\limits_{1 \leq i < j \leq k+1}^{k+1}P(A_iA_j) + \cdots +(-1)^{k+2}P(\bigcap\limits_{i=1}^{k+1}A_i) \)
符合原命题结构。
因此，原命题成立.