As a mathematician with a deep understanding of number theory, I can provide a comprehensive explanation of the properties of odd and even numbers and address the specific question you've asked: "Is \(2n + 1\) always odd?" To begin with, let's define what even and odd numbers are. An even number is any integer that can be expressed as \(2k\), where \(k\) is an integer. This means that when you divide an even number by 2, there is no remainder. On the other hand, an odd number is an integer that can be expressed as \(2k + 1\), where \(k\) is also an integer. Dividing an odd number by 2 always leaves a remainder of 1. Now, let's consider the expression \(2n + 1\). Here, \(n\) is any integer, which means it could be even or odd. If \(n\) is even, say \(n = 2k\), then \(2n + 1 = 2(2k) + 1 = 4k + 1\). Since \(4k\) is clearly even (as it is a multiple of 2), adding 1 to it will result in an odd number because it will leave a remainder of 1 when divided by 2. If \(n\) is odd, say \(n = 2k + 1\), then \(2n + 1 = 2(2k + 1) + 1 = 4k + 2 + 1 = 4k + 3\). Here, \(4k\) is again even, and \(4k + 3\) is odd for the same reason as before. The key insight here is that the expression \(2n + 1\) is structured in such a way that it will always result in an odd number, regardless of whether \(n\) is even or odd. This is because the term \(2n\) will always be even (since it's a multiple of 2), and adding 1 to an even number always results in an odd number. The reference content you provided also touches on this concept. It states that if \(n\) is any integer, \(2n\) will always be even because it is a multiple of 2. The expression \(2n(n + n)\) simplifies to \(2n^2\), which is also even for the same reason. The integers \(2n + 1\) and \(2n - 1\) are indeed the preceding and succeeding integers of \(2n\), and they are always odd because they are one more or one less than an even number, respectively. In conclusion, the expression \(2n + 1\) is always odd, no matter what integer value \(n\) takes. This is a fundamental property of integers and their parity (even or odd). Understanding these properties is crucial for anyone studying mathematics, as they form the basis for more advanced topics in number theory and algebra. read more >>

If n is any integer (both even and odd) is added to itself always generates an even number. Hence 2n(n+n) is always an even number. The integers 2n+1 and 2n-1 are preceding and succeeding integers of 2n, hence the are always odd numbers.read more >>