- #1
mathstudent88
- 27
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If n is a natural number then n2+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 n2+n+3
=(2k)2+2k+3
=4k2+2k+3
= 2(2k2+k+1)+1, where (2k2+k+1) is a member of the natural numers. This means that when n is even, n2+n+3 is odd.
If n is odd, then n=2j+1 where j is a member of the natural numbers
and n2+n+3
=(2j+1)2+(2j+1)+3
=4j2+4j+1+2j+4
=4j2+6j+5
=2(2j2+3j+2)+1, where (2j2+3j+2) is a memeber of the natural numbers. This means that when n is odd, n2+n+3.
Is this ok? Thanks for the help!
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 n2+n+3
=(2k)2+2k+3
=4k2+2k+3
= 2(2k2+k+1)+1, where (2k2+k+1) is a member of the natural numers. This means that when n is even, n2+n+3 is odd.
If n is odd, then n=2j+1 where j is a member of the natural numbers
and n2+n+3
=(2j+1)2+(2j+1)+3
=4j2+4j+1+2j+4
=4j2+6j+5
=2(2j2+3j+2)+1, where (2j2+3j+2) is a memeber of the natural numbers. This means that when n is odd, n2+n+3.
Is this ok? Thanks for the help!