(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For n ∈ Z+ prove by contrapositive that if 2n3 + 3n2 + 4n + 5 is odd then n is even.

2. Relevant equations

3. The attempt at a solution

If n is odd then ( 2n³ + 3n² + 4n + 5 ) is even.

By definition, a number n is odd if n = 2k + 1 for some integer k. A number n is even if n = 2k for some integer k. If x and y are two integers for which x + y is even, then by definition x and y have the same parity. It follows that two integers will have the same parity if they are both odd or both even.

2(2k + 1)³ + 3(2k + 1)² + 4(2k + 1) + 5

We see here that 2(2k + 1)³ can be written as 2n where n = (2k+1)³. Thus we conclude 2(2k + 1)³ is an even number.

By the same logic, 4(2k + 1) can be written as 2n where n = 2(2k+1).

I get stuck after this... How can i prove 3(2k + 1)² is odd so that i can add it to 5 and get even parity?

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# Homework Help: [Proofs] Proof by contraposition

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