Summation of i from 1 to n of Summation of j from 1 to i/Proof 2
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Theorem
- $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j} = \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}$
Proof
\(\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i a_{i j}\) | \(=\) | \(\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le i \le n}\right] \left[{1 \le j \le i}\right]\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le i \le n}\right]\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i, j \mathop \in \Z} a_{i j} \left[{1 \le j \le n}\right] \left[{j \le i \le n}\right]\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{j \mathop = 1}^n \sum_{i \mathop = j}^n a_{i j}\) |
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products