(adsbygoogle = window.adsbygoogle || []).push({}); If n is a natural number then n^{2}+n+3 is odd.

This is what I have and wanted to know if I was doing it right or not:

Let n be a member of the natural numbers. If n is even, then n=2k, k member of natural numbers, and n^{2}+n+3

=(2k)^{2}+2k+3

=4k^{2}+2k+3

= 2(2k^{2}+k+1)+1, where (2k^{2}+k+1) is a member of the natural numers. This means that when n is even, n^{2}+n+3 is odd.

If n is odd, then n=2j+1 where j is a member of the natural numbers

and n^{2}+n+3

=(2j+1)^{2}+(2j+1)+3

=4j^{2}+4j+1+2j+4

=4j^{2}+6j+5

=2(2j^{2}+3j+2)+1, where (2j^{2}+3j+2) is a memeber of the natural numbers. This means that when n is odd, n^{2}+n+3.

Is this ok? Thanks for the help!

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# Natural numbers proof help

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