prove that if n is a positive integer, then 4 divides 1+(-1)^n (2n-1)

To prove that \( 4 \) divides \( 1 + (-1)^n (2n - 1) \) for any positive integer \( n \), we need to show that:

\[
1 + (-1)^n (2n - 1) \equiv 0 \pmod{4}
\]

We'll analyze this expression in two cases: one for \( n \) being even and one for \( n \) being odd.

### Case 1: \( n \) is even

Let \( n = 2k \), where \( k \) is a non-negative integer. Substituting \( n = 2k \) into the expression:

\[
1 + (-1)^{2k} (2(2k) - 1)
\]

Since \( (-1)^{2k} = 1 \) for any integer \( k \), this becomes:

\[
1 + 1 (4k - 1) = 1 + 4k - 1 = 4k
\]

Clearly, \( 4k \) is divisible by 4, so in this case, \( 1 + (-1)^n (2n - 1) \equiv 0 \pmod{4} \).

### Case 2: \( n \) is odd

Let \( n = 2k + 1 \), where \( k \) is a non-negative integer. Substituting \( n = 2k + 1 \) into the expression:

\[
1 + (-1)^{2k+1} (2(2k+1) - 1)
\]

Since \( (-1)^{2k+1} = -1 \), this becomes:

\[
1 + (-1) (4k + 1) = 1 - (4k + 1) = 1 - 4k - 1 = -4k
\]

Again, \( -4k \) is divisible by 4, so in this case as well, \( 1 + (-1)^n (2n - 1) \equiv 0 \pmod{4} \).

### Conclusion:
In both cases (whether \( n \) is even or odd), the expression \( 1 + (-1)^n (2n - 1) \) is divisible by 4. Thus, we have proven that:

\[
4 \text{ divides } 1 + (-1)^n (2n - 1) \text{ for all positive integers } n.
\]

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### Related Questions:
1. What happens if we generalize this to expressions of the form \( 1 + (-1)^n (2n - k) \)?
2. Can we prove a similar divisibility property for other values of \( k \)?
3. How would the proof change if we considered \( n \) as a negative integer?
4. What divisibility properties can we infer for more complex expressions involving powers of \( (-1)^n \)?
5. How can we use modular arithmetic to simplify similar proofs?

**Tip**: Breaking problems into cases (like odd/even \( n \)) is a powerful technique in many proofs!

Prove that if n is a positive integer, then 4 divides 1 + (-1)^n (2n - 1).

Explore the proof that for any positive integer n, 4 divides 1 + (-1)^n (2n - 1). This solution breaks down the problem using modular arithmetic and case analysis (even and odd n), covering divisibility properties and powers of -1. Suitable for high school students in Grades 10-12.

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