Summation of i from 1 to n of Summation of j from 1 to i/Proof 2

Theorem

$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$

$\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j}$

$=$

$\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le i \le n}\right] \left[{1 \le j \le i}\right]$

$\ds $

$=$

$\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le i \le n}\right]$

$\ds $

$=$

$\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le n}\right] \left[{j \le i \le n}\right]$

$\ds $

$=$

$\ds \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$

$\blacksquare$