prove the following statement by cases for any integer n show that n^2 + 15n - 1 is odd so we will consider 2 cases n is even and n is odd in this step you will complete the proof of the first case so case 1: n is even let n be an even integer by the definition of even BLANK for some integer k this follows that n^2 + 15n - 1= 2*(BLANK) + 1 since BLANK is an integer BLANK must be an integer since n^2 + 15n - 1 equal BLANK times an integer plus BLANK, n^2 + 15n - 1 is BLANK so just gimme the answers of the blanks

Let's fill in the blanks step by step for **Case 1: n is even**:

1. **Blank 1**: \( n = 2k \) (By the definition of even, \( n \) is a multiple of 2 for some integer \( k \).)
   
2. **Blank 2**: \( k^2 + 8k \) (This is the expression you get when simplifying \( n^2 + 15n - 1 \) in terms of \( k \).)

3. **Blank 3**: \( k^2 + 8k \) (Again, this refers to the expression \( k^2 + 8k \) as part of the simplification process.)

4. **Blank 4**: \( k^2 + 8k \) (This is the same expression \( k^2 + 8k \) being referred to, which is an integer.)

5. **Blank 5**: odd (Since \( n^2 + 15n - 1 = 2 \times \) an integer \( + 1 \), it must be odd.)

So the full proof in this case should look like:
- Let \( n \) be an even integer. By the definition of even, \( n = 2k \) for some integer \( k \).
- Substituting into the expression \( n^2 + 15n - 1 \), we get \( (2k)^2 + 15(2k) - 1 = 4k^2 + 30k - 1 = 2(k^2 + 8k) + 1 \).
- Since \( k^2 + 8k \) is an integer, the expression \( n^2 + 15n - 1 \) equals \( 2 \times \) an integer \( + 1 \), so it must be odd.

Would you like more details on this proof?

Here are 5 related questions you might be interested in:
1. How do you prove the second case when \( n \) is odd?
2. What general strategy can we use to prove statements by cases?
3. How do odd and even numbers interact under different mathematical operations?
4. Can this method be applied to other types of quadratic expressions?
5. What are some common properties of odd and even functions in mathematics?

**Tip**: When working with even and odd numbers in algebra, always look for patterns that arise from using \( 2k \) for even and \( 2k + 1 \) for odd integers.

Prove the following statement by cases: For any integer n, show that n^2 + 15n - 1 is odd. We will consider two cases: n is even and n is odd. In this step, complete the proof of the first case. Case 1: n is even. Let n be an even integer. By the definition of even BLANK for some integer k. This follows that n^2 + 15n - 1 = 2*(BLANK) + 1. Since BLANK is an integer, BLANK must be an integer. Since n^2 + 15n - 1 equals BLANK times an integer plus BLANK, n^2 + 15n - 1 is BLANK.

Explore the proof that n^2 + 15n - 1 is odd for any integer n, using a case-by-case approach for both even and odd numbers. This solution breaks down the proof for when n is even, demonstrating the use of number theory and quadratic expressions. Suitable for students in Grades 9-11.

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