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Math Help - a proof

  1. #1
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    a proof

    Use mathematical induction to prove that 1^2 + 2^2 + 3^2.... + n^2 = (n(n+1)(2n+1))/6

    thanks!
    Last edited by s0urgrapes; July 26th 2009 at 02:39 PM.
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  2. #2
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    Quote Originally Posted by s0urgrapes View Post
    Use mathematical induction to prove that 1^2, 2^2, 3^2....n^2 = (n(n+1)(2n+1))/6
    It should be 1^2+ 2^2+ 3^2+....+n^2 = (n(n+1)(2n+1))/6
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  3. #3
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    It's obviously true for n=1.

    Assume  1^{2}+2^{2}+3^{2}+k^{2} = \frac {k(k+1)(2k+1)}{6} is true for some k>1.

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

     = \frac{2k^{3}+9k^{2}+13k+6}{6}

     = \frac{(k+1)(2k^{2}+7k+6)}{6}

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

     = \frac{\big(k+1\big)\big((k+1)+1\big)\big(2(k+1)+1\  big)}{6}
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