Prove by contradiction that if n is odd, n^(3)+1 is even.

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Prove by contradiction that if  n is odd,  n^(3)+1 is even.

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Assume n is odd: Assume for the sake of contradiction that n is odd and n^3 + 1 is not even, which means n^3 + 1 is odd.Express n as 2k + 1: An odd number can be expressed in the form of 2k + 1, where k is an integer. Since n is odd, let n = 2k + 1.Substitute n in expression: Substitute n with 2k + 1 in the expression n^3 + 1 to find its value in terms of k. n^3 + 1 = (2k + 1)^3 + 1Expand (2k + 1)^3: Expand the cube (2k + 1)^3 using the binomial theorem or by multiplying (2k + 1)(2k + 1)(2k + 1). (2k + 1)^3 = 2k(2k(2k + 1) + 1) + 1 = 8k^3 + 12k^2 + 6k + 1Add 1 to get n^3 + 1: Add 1 to the expanded form of (2k + 1)^3 to get n^3 + 1. n^3 + 1 = 8k^3 + 12k^2 + 6k + 1 + 1 = 8k^3 + 12k^2 + 6k + 2Identify even components: Notice that 8k^3, 12k^2, and 6k are all multiples of 2, which means they are even. Adding even numbers together will result in an even number. Adding 2 to an even number will still result in an even number.Contradiction of initial assumption: Since n^3 + 1 = 8k^3 + 12k^2 + 6k + 2 is even, our initial assumption that n^3 + 1 is odd leads to a contradiction.Conclusion by contradiction: Therefore, by contradiction, if n is odd, then n^3 + 1 must be even.

Prove by contradiction that if $n$ is odd, $n^{3}+1$ is even.

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