Summation of i from 1 to n of Summation of j from 1 to i/Proof 1
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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
- $\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^i$
can be expressed as:
- $\ds \sum_{\map R i} \sum_{\map S {i, j} } a_{i j}$
where:
- $\map R i$ is the propositional function $1 \le i \le n$
- $\map S {i, j}$ is the propositional function $1 \le j \le i$
We wish to find a propositional function $\map {S'} j$ which is to be:
- there exists an $i$ such that both $1 \le i \le n$ and $1 \le j \le i$
This is satisfied by the propositional function:
- $\map {S'} j := 1 \le j \le n$
Next we wish to find a propositional function $\map {R'} {i, j}$ which is to be:
- both $1 \le i \le n$ and $1 \le j \le i$
This is satisfied by the propositional function:
- $\map {R'} {i, j} := j \le i \le n$
Hence the result, from Exchange of Order of Summation with Dependency on Both Indices.
$\blacksquare$
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: $(10)$