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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    I'm learning proof by induction at the moment and having some difficulties understanding. Taking the wikipedia entry for it, can someone explain the final step:

    Induction Example

    (0 + 1 + 2 .... + k + k + 1) = \frac{(k+1)((k+1) + 1)}{2}

    Using induction Hypothesis that P(k) holds, the left side can be rewritten:

    \frac{k(k+1)}{2} +(k+1)

    It's the next part I don't get:

    Algebraically:

    \frac{k(k+1)}{2} +(k+1) = \frac{k(k+1)+2(k+1)}{2}

    \frac{k(k+1)}{2} +(k+1) = \frac{(k+1)(k+2)}{2}

    \frac{k(k+1)}{2} +(k+1) = \frac{(k+1)((k+1) + 1)}{2}

    Where are the first two expressions coming from on the right hand side here? :?

    Cheers

    BIOS
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  2. #2
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    Re: Proof by Induction

    Quote Originally Posted by BIOS View Post
    (0 + 1 + 2 .... + k + k + 1) = \frac{(k+1)((k+1) + 1)}{2}
    Using induction Hypothesis that P(k) holds, the left side can be rewritten:
    \frac{k(k+1)}{2} +(k+1)

    It's the next part I don't get:

    Algebraically:

    \color{blue}\frac{k(k+1)}{2} +(k+1) = \frac{k(k+1)+2(k+1)}{2}

    \color{blue}\frac{k(k+1)}{2} +(k+1) = \frac{(k+1)(k+2)}{2}

    \frac{k(k+1)}{2} +(k+1) = \frac{(k+1)((k+1) + 1)}{2}
    Where are the first two expressions coming from on the right hand side here? :?
    Are you asking about those two expressions?
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  3. #3
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    Re: Proof by Induction

    Yeah. Sorry didn't know how to highlight! Specifically the right hand side.
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  4. #4
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    Re: Proof by Induction

    Quote Originally Posted by BIOS View Post
    Yeah. Sorry didn't know how to highlight! Specifically the right hand side.
    In the first one the common denominator is 2.

    The in the second factor out the (k+1)
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  5. #5
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    Re: Proof by Induction

    Gotcha! Thanks for the reply as always

    Cheers

    BIOS
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