Question

Prove the following statement by cases: for any integer n, show that n² + 5n – 1 is odd. We will prove by two cases: n is even and n is odd. In this step, you will complete Proof. Case 1: Let n be an even integer. By the definition of even, there exists an integer k such that n = 2K This follows that na + 5n -1 = (2k)2 + 5(2k) – 1 = 2.( )+1. Since k is an integer, and integers are closed under and must be an integer.By the definition of odd is .

Answer 1

The expression (n² + 5n - 1) is always an odd integer for any integer value of (n).

Step-by-step explanation:

To prove this statement, we'll consider two cases: when (n) is even and when (n) is odd.

(n) is even:

Let (n = 2k) for some integer (k). Substituting this into the expression (n² + 5n - 1) gives us:

[n² + 5n - 1 = (2k)² + 5(2k) - 1 = 4k² + 10k - 1]

Now, consider the expression (4k² + 10k - 1) where both (4k² ) and (10k) are even integers (since they have a factor of 2). When we subtract 1 from an even number, the result is always odd. Hence, (4k² + 10k - 1) is an odd integer for any integer (k).

(n) is odd:

Let (n = 2k + 1) for some integer (k). Substituting this into the expression (n² + 5n - 1) gives us:

[n² + 5n - 1 = (2k + 1)² + 5(2k + 1) - 1 = 4k² + 4k + 1 + 10k + 5 - 1 = 4k² + 14k + 5]

The expression (4k² + 14k + 5) can be rewritten as (2(2k² + 7k) + 5), where (2k² + 7k) is an integer. When we add 5 to an even number, the result is always odd. Hence, (4k² + 14k + 5) is an odd integer for any integer (k).

Therefore, we've shown that whether (n) is even or odd, the expression (n² + 5n - 1) always evaluates to an odd integer, thus proving the initial statement true for all integer values of (n).