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    Hi, I have a homework question:

    Show that \displaystyle\sum_{r=n+1}^{2n} r^2= \frac{1}{4}n^2(3n+1)(5n+3)

    I split it into:

    \displaystyle\sum_{r=n+1}^{2n} r^2 = \displaystyle\sum_{r=1}^{2n} r^2  - \sum_{r=1}^{n} r^2

    I know what the second bit is but was wondering what \displaystyle\sum_{r=1}^{2n} r^2 is.

    Many thanks in advance
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    (Original post by bendent1234)
    Hi, I have a homework question:

    Show that \displaystyle\sum_{r=n+1}^{2n} r^2= \frac{1}{4}n^2(3n+1)(5n+3)

    I split it into:

    \displaystyle\sum_{r=n+1}^{2n} r^2 = \displaystyle\sum_{r=1}^{2n} r^2  - \sum_{r=1}^{n} r^2

    I know what the second bit is but was wondering what \displaystyle\sum_{r=1}^{2n} r^2 is.

    Many thanks in advance
    You know the formula for the sum from 1 to n, and it's expressed as a function of n.

    If you want the sum from 1 to m, the formula would be the same, just with n replaced by m.

    Similarly, for the sum from 1 to 2n, again the basic formula is the same, but just replace the n with 2n.
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    (Original post by ghostwalker)
    You know the formula for the sum from 1 to n, and it's expressed as a function of n.

    If you want the sum from 1 to m, the formula would be the same, just with n replaced by m.

    Similarly, for the sum from 1 to 2n, again the basic formula is the same, but just replace the n with 2n.
    Thanks, I thought it might be something like that but thanks for clarifying
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    (Original post by bendent1234)
    Thanks, I thought it might be something like that but thanks for clarifying
    You're welcome, and +rep for a well laid out question.
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    When subbing 2n into the formula to get that the answer is:

    \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}

    I got this to be:

    \frac{n(7n+1)(2n+1)}{6}

    This is clearly not what it should have been but I am not sure where I went wrong as my workings all seem mathematically correct
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    (Original post by bendent1234)
    When subbing 2n into the formula to get that the answer is:

    \frac{2n(2n+1)(4n+1)}{6} - \frac{n(n+1)(2n+1)}{6}

    I got this to be:

    \frac{n(7n+1)(2n+1)}{6}

    This is clearly not what it should have been but I am not sure where I went wrong as my workings all seem mathematically correct
    Your answer is correct. The question is wrong - the question should be the sum of the cube of r.
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    (Original post by Mr M)
    Your answer is correct. The question is wrong - the question should be the sum of the cube of r.
    Thanks for clarifying that, my teacher probably copied it onto the board wrong. I shall do the correct question
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    (Original post by bendent1234)
    Thanks for clarifying that, my teacher probably copied it onto the board wrong. I shall do the correct question
    I don't know if this helps for catching things like this, but the "leading term" in a summation should be the same as the leading term in what you got if you integrated instead.

    e.g.

    \sum_1^n r^3 = \frac{1}{4}n^2(n+1)^2 = \frac{n^4}{4} + {terms smaller than n^4}. While \int_1^n r^3 = \frac{n^4}{4} - \frac{1}{4}.

    This works even if you have unusual upper/lower limits as in the original question.
 
 
 
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