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Math Help - elem. numb. theory - hexagonal number nth formula - proof

  1. #1
    Member cassiopeia1289's Avatar
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    elem. numb. theory - hexagonal number nth formula - proof

    OK: so, I found that the hexagonal number for nth term is:

    1+5+9+...+(4n-3) = n(2n-1)

    Proof:
    (Eventually, I will want to get to 1+5+...+(4(k+1)-3) = (k+1)(2(k+1)-1) = 2k^2 + 3k + 1

    SO:
    Let P(n) be the statement 1+5+...+(4n-3) = n(2n-1).
    (Base case) Since 4(1)-3 = 1, which equals (1)(2(1)-1), P(1) is true.
    Let k be a pos. integer.
    So 1+5+...+(4k-3) = k(2k-1)
    Therefore, 1+5+...+(4k-3) + (4(k+1) - 3) = k(2k-1) + (4(k+1) - 3)
    = 2k^2 - k + 4k + 3
    = 2k^2 + 3k + 3

    Did i do something wrong? I have an extra +2 in the mix.
    Please work with what I have, if possible, I kinda need to use the Strong Induction.
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  2. #2
    o_O
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    k(2k-1) + 4(k+1) - 3

    = 2k^2 - k + 4k + {\color{red}4} - 3

    = 2k^2 + 3k + {\color{red}1}

    Just a little arithmetic error
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  3. #3
    Member cassiopeia1289's Avatar
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    aw thank you! I always end up doing stupid stuff like that
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