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Math Help - Formula in terms of n for the sum

  1. #1
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    Smile Formula in terms of n for the sum

    I want to say thank you to plato because I think I an understand what the book is asking. This problem asks to find a formula in terms of n for the sum where the lower limit in sigma notation is k=1 the upper limit=n and k=sum of the integers. So I was thinking that the lower limit is k=1 and the upper limit is n then the sum of the integers K = n(n+1)/2, is this correct? and is the identity of this formula: (n+1)^3-1^3 ?
    Please let me know if I am on the right track, I would appreciate any feedback you could give me.

    Thank you!!
    Keith Stevens
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kcsteven View Post
    I want to say thank you to plato because I think I an understand what the book is asking. This problem asks to find a formula in terms of n for the sum where the lower limit in sigma notation is k=1 the upper limit=n and k=sum of the integers. So I was thinking that the lower limit is k=1 and the upper limit is n then the sum of the integers K = n(n+1)/2, is this correct? and is the identity of this formula: (n+1)^3-1^3 ?
    Please let me know if I am on the right track, I would appreciate any feedback you could give me.

    Thank you!!
    Keith Stevens
    Yes, \sum_{k = 1}^n = \frac{n(n+1)}{2}

    What do you mean by "the identity of this formula"?

    -Dan
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    Forum Admin topsquark's Avatar
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    Copied from a PM from kcsteven:

    I did not proof read the whole question, and it does not make sense. I was asking if I used the formula expressed in sigma where the lower limit is k=a and the upper limit is n and then n(n+1)/2, I could use the formula and the identity (n+1)^3 -n^3 = 3n^2 + 3n +1, to find a formula in sigma where the lower limit is k=1 the upper is n and k^2 is the sum of the squares of the integers. would it be n(n+1)(2n+1)/6, If this is correct what would be a good example?
    If I'm wrong and all you are asking is verification, then yes:
    \sum_{k = 1}^n k^2 = \frac{n(n+1)(2n+1)}{6}

    If you want to know how to derive this, then this is a job for Soroban, who does an unusually excellent job of describing how to do this.

    -Dan
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