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Math Help - Prove theory

  1. #1
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    Prove theory

    Hi Guys,

    Preparing for final and came across home work question.

    Question: Prove that n^2 + n is even for any interger n.

    Please somebody help urgently.

    Thanks
    CHicago boy
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  2. #2
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    Hello.

    Quote Originally Posted by chicago_boy81 View Post
    Hi Guys,

    Preparing for final and came across home work question.

    Question: Prove that n^2 + n is even for any interger n.
    Two cases

    1) n is odd: then n = 2m+1, m\in \mathbb{N}_0

    Thus
     n^2 + n = (2m+1)^2 + 2m+1 =(4m^2+2+1)+2m+1 = 4m^2+4+2m = 2(2m^2+4+2m)

    is even

    2) n is even: then n = 2m, m \in \mathbb{N}

    => n^2+n = (2m)^2 + 2m = 4m^2 + 2m = 2(2m^2+m)

    Do you understand?

    Regards,
    Rapha
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  3. #3
    MHF Contributor chisigma's Avatar
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    Very easy...

    ... for n even n and n^{2} are both even, so that n^{2} + n is even...


    ... for n odd n and n^{2} are both odd, so that n^{2} + n is even...

    Kind regards

    \chi \sigma
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  4. #4
    MHF Contributor Swlabr's Avatar
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    A third solution won't harm anyone:

    n^2+n = n(n+1) and clearly 2 divides one of n or n+1.
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  5. #5
    MHF Contributor chisigma's Avatar
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    We can conclude remembering the 'legend' according to which Gauss demonstrated when he was seven years old [!] that...

    \sum_{i=1}^{n} i = \frac {n\cdot (n+1)}{2} (1)

    True or not the 'legend', it is evident from (1) that n\cdot (n+1) must be an even number...

    Kind regards

    \chi \sigma
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  6. #6
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    It looks like your proof is going to 2 cases, the 1st case being if n is even so n=2k for some integer k, and the second case will be for if n is odd so n=2m+1 for some integer m. This covers the set all of integers because an integer is either odd or even. hope this helps =).
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