Sum of Even Number of Odd Numbers is Even

Proof
Let $S = \set {r_1, r_2, \ldots, r_n}$ be a set of $n$ odd numbers, where $n = 2 m$.

By definition of odd number, this can be expressed as:
 * $S = \set {2 s_1 + 1, 2 s_2 + 1, \ldots, 2 s_n + 1}$

where:
 * $\forall k \in \closedint 1 n: r_k = 2 s_k + 1$

Then:

Thus, by definition, $\ds \sum_{k \mathop = 1}^n r_k$ is even.