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An odd integer is a whole number that cannot be divided evenly by 2. In other words, if you divide an odd number by 2, you'll always have a remainder of 1. Examples of odd integers include numbers like 1, 3, 5, 7, and so forth.

To express odd integers mathematically, we often use the formula \(2n + 1\), where \(n\) is an integer. This formula neatly captures the idea that when you take any integer \(n\) and multiply it by 2 (resulting in an even number), adding 1 will always make the result odd. This property of odd numbers is key to understanding how they behave in addition and other operations. Remember, the sum or difference of two odd integers will always be even, but that changes when we have an odd number of odd integers involved, as shown in the problem solution.

A mathematical proof is a logical argument that demonstrates the truth of a given statement or proposition. It uses established mathematical principles and logical steps to arrive at a conclusion. Proofs are fundamental to mathematics because they help us establish facts beyond doubt.

When approaching a proof, it often starts with defining the objects involved, like how odd integers were defined in the exercise. Then, through a series of logical steps, the goal is to show that one statement leads to another, eventually reaching the conclusion. In this exercise, each step builds on the previous one, proving that the sum of an odd number of odd integers remains odd by manipulating the expressions used in the proof. A successful proof can have broader implications, confirming similar truths in other areas of mathematics as well.

When we talk about the summation of odd numbers, it usually means adding several odd integers together. This activity can be particularly interesting because of the unique properties of odd numbers.

In the given exercise, the goal was to show that the sum of an odd number of odd integers is itself odd. To understand why, consider the sum \(S = (2n_1 + 1) + (2n_2 + 1) + \, ... \, + (2n_k + 1)\), which resulted in \(S = 2m + k\) after simplifying. Here, \(2m\) is an even number since it’s 2 times another number \(m\), and \(k\) is odd given the problem statement.

When you add an odd number to an even number, the result is always odd. This is because the extra 1 from the odd number doesn’t "pair" up, leaving a remainder of 1 when divided by 2. This feature is demonstrated in our proof and highlights the fascinating consistency in the behavior of odd numbers during summation.