Product of Summations

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Theorem

$\ds \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j} = \sum_{1 \mathop \le i_1 \mathop , \mathop \ldots \mathop , i_n \mathop \le m} a_{i_1 1} \cdots a_{i_n n}$


Proof

\(\ds \prod_{j \mathop = 1}^n \sum_{i \mathop = 1}^m a_{i j}\) \(=\) \(\ds \sum_{i \mathop = 1}^m a_{i 1} \sum_{i \mathop = 1}^m a_{i 2} \cdots \sum_{i \mathop = 1}^m a_{i n}\) Definition of Continued Product
\(\ds \) \(=\) \(\ds \sum_{i_1 \mathop = 1}^m a_{i_1 1} \sum_{i_2 \mathop = 1}^m a_{i_2 2} \cdots \sum_{i_n \mathop = 1}^m a_{i_n n}\)
\(\ds \) \(=\) \(\ds \sum_{i_1 \mathop = 1}^m \sum_{i_2 \mathop = 1}^m \cdots \sum_{i_n \mathop = 1}^m a_{i_1 1} a_{i_2 2} \cdots a_{i_n n}\) General Distributivity Theorem
\(\ds \) \(=\) \(\ds \sum_{1 \mathop \le i_1 \mathop , \mathop \ldots \mathop , i_n \mathop \le m} a_{i_1 1} \cdots a_{i_n n}\)

$\blacksquare$


Sources

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