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Math Help - Summation identity

  1. #1
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    Summation identity

    Why is the following identity true?

    \sum_{r=1}^{n}r^2 = \frac{1}{6} n(n+1)(2n+1)
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Summation identity

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  3. #3
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    Re: Summation identity

    sorry, had problems posting the equation. It is there now
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  4. #4
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    Re: Summation identity

    The best way to answer this 'why?' is to prove it yourself.

    Do you know of mathematical induction?
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    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Summation identity

    Quote Originally Posted by zhangvict View Post
    Why is the following identity true?

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

    \sum_{r=0}^{n}r^2=?

    Solution via 'Differences Series' (Or LaTex practice )


    Step no. 0:
    \sum_{r=0}^{n}r^2

    Step no. 1:
    \sum_{r=0}^{n+1}r^2-\sum_{r=0}^{n}r^2=n^2

    Step no. 2:
    n^2-(n-1)^2=2n-1

    Step no. 3:
    2n-1-[2(n-1)-1]=2

    Step no. 4:
    2-2=0

    Now we find the generating function from 4 and up...

    The generating function of #4 is x+x^2

    The generating function of #3 is \frac{x+x^2}{1-x}

    The generating function of #2 is \frac{x+x^2}{(1-x)^2}

    The generating function of #1 is \frac{x+x^2}{(1-x)^3}

    The generating function of #0 is \frac{x+x^2}{(1-x)^4}

    We get that generating function of (\sum_{r=0}^{n}r^2)_{n=0}^{\infty} is \frac{x+x^2}{(1-x)^4}.

    So, \sum_{r=0}^{n}r^2 is the coefficient of x^n in \frac{x+x^2}{(1-x)^4}:

    \frac{x+x^2}{(1-x)^4}=(x+x^2)\cdot \frac{1}{(1-x)^4}=(x+x^2)\sum_{i=0}^{\infty}\binom{i+3}{3}x^i

    Hence the coefficient of x^n is:

    \binom{n-1+3}{3}+\binom{n-2+3}{3}

    The conclusion is:

    \sum_{r=0}^{n}r^2=\binom{n+2}{3}+\binom{n+1}{3}
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  6. #6
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    Re: Summation identity

    Quote Originally Posted by zhangvict View Post
    Why is the following identity true?

    \sum_{r=1}^{n}r^2 = \frac{1}{6} n(n+1)(2n+1)
    There are a number of ways to show why the formula holds.

    One easy method is to use the fact that there is a relationship
    between the sum of positive natural numbers
    and the sum of their squares....

    \sum_{r=1}^nr=\frac{n(n+1)}{2}

    as it is a very basic arithmetic series.

    1+2=3;\;\;\;1^2+2^2=3\left(\frac{5}{3}\right)

    1+2+3=6;\;\;\;1^2+2^2+3^2=6\left(\frac{7}{3}\right  )

    1+2+3+4=10;\;\;\;\;1^2+2^2+3^2+4^2=10\left(\frac{9  }{3}\right)

    Hence, induction can prove without doubt that

    \sum_{r=1}^nr^2=\frac{2n+1}{3}\sum_{r=1}^nr
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  7. #7
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    Re: Summation identity



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