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Math Help - [SOLVED] Please help with proof

  1. #1
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    [SOLVED] Please help with proof

    I have no idea how to go about it really. Any help is very much appreciated.

    <br />
\sum_{k=1}^{n}k^2 = \frac{n(n+1)(2n+1)}{6}<br />
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  2. #2
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    Hi nunos

    I suggest using mathematical induction
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  3. #3
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    Quote Originally Posted by songoku View Post
    Hi nunos

    I suggest using mathematical induction
    I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

    Proving for n = 1, easy.

    Assuming it's true for n, let's evaluate for n+1.

    \sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}

    \sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)

    And I don't know how to go from here... I don't even know if what I've done is relevant...
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  4. #4
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    Quote Originally Posted by nunos View Post
    I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

    Proving for n = 1, easy.

    Assuming it's true for n, let's evaluate for n+1.

    \sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}

    \sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)

    And I don't know how to go from here... I don't even know if what I've done is relevant...
    I believe you have some errors here. I think what you want is

    \sum_{k=1}^{n+1} k^2 =\sum_{k=1}^{n} k^2 +(n+1)^2 = \frac{n(n+1)(2n+1)}{6} + (n+1)^2 = \frac{(n+1)(n+2)(2n+3)}{6}
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  5. #5
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    Quote Originally Posted by nunos View Post
    I already know about that, but thanks for the suggestion anyway. I am going to write here, what I have so far, and hopefully you or someone will point me in the right direction.

    Proving for n = 1, easy.

    Assuming it's true for n, let's evaluate for n+1.

    \sum_{k=1}^{n+1} k^{n+1} = \sum_{k=1}^n k^n + k^{n+1} = \frac{n(n+1)(2n+1)}{6} + k^{n+1}
    No, not k^{n+1}! Your sum was \sum_{k=1}^n k^2. You want \sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (k+1)^2.

    \sum_{k=1}^{n+1} k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Rightarrow \frac{n(n+1)(2n+1)}{6} + k^{n+1} = \frac { (n+1)(n+2)(2n+3)}{6}

    \Leftrightarrow n(n+1)(2n+1) + 6k^{n+1} = (n+1)(n+2)(2n+3)

    And I don't know how to go from here... I don't even know if what I've done is relevant...
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    No, not k^{n+1}! Your sum was \sum_{k=1}^n k^2. You want \sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (k+1)^2.
    Just a small correction here (just want to make sure the OP isn't confused).

    Should be:

    \sum_{k=1}^{n+1} k^2= \sum_{k=1}^n k^2+ (\textcolor{red}{n}+1)^2
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  7. #7
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    You're all correct. That was a mistake. I had the impression that something had to be wrong. I had now completed the proof.

    Thanks.
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