Final answer:
The theorem is proven by contraposition by assuming n is odd and showing that (n+1)^2 results in an even number. The proof by contradiction assumes both n and (n+1)^2 are odd, which leads to a contradiction, thereby confirming the theorem.
Step-by-step explanation:
Proof by Contraposition
To prove the theorem 'If (n+1)^2 is odd, then n is even' by contraposition, we start with the contrapositive statement: 'If n is not even (i.e., n is odd), then (n+1)^2 is not odd (i.e., (n+1)^2 is even)'. Since n is odd, we can express n as 2k+1, where k is an integer. Then, (n+1)^2 becomes [(2k+1)+1]^2 = (2k+2)^2 = (2(k+1))^2. The last expression represents the square of an even number (since it is 2 times another integer), which is always even. Thus, if n is odd, (n+1)^2 is indeed even, which confirms the contrapositive and therefore proves the original theorem.
Proof by Contradiction
To prove the theorem via contradiction, let's assume the opposite of what we want to prove: 'Assume that (n+1)^2 is odd and n is odd'. Let n be an odd integer, so n=2k+1 for an integer k. Then (n+1)^2 = [2k+1+1]^2 = (2k+2)^2 = 4(k+1)^2, which is clearly a multiple of 4 and, hence, even. This contradicts our assumption that (n+1)^2 is odd. Therefore, our initial assumption must be wrong and it must be that if (n+1)^2 is odd, then n is even