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Math Help - Sequence and series help

  1. #1
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    Sequence and series help

    how would i find the function? The series is 1,2,2,3,3,3,4,4,4,4 and so on with there being n number of the term n. How would i make a function to find the nth term?

    Grateful for help.
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  2. #2
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    You can try

    u_n = -E(\frac{1-\sqrt{8n+1}}{2})

    where k \leq E(k) < k+1

    If needed I can provide you the way I found this function
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  3. #3
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    How?

    Yes, I want to know how you got it.
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  4. #4
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    Quote Originally Posted by sethu1989 View Post
    Yes, I want to know how you got it.
    C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil
    That is the ceiling function.
    The 'change' come on the triangular numbers: \frac {n(n+1)}{2}.
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  5. #5
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    Quote Originally Posted by Plato View Post
    C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil
    That is the ceiling function.
    The 'change' come on the triangular numbers: \frac {n(n+1)}{2}.
    Thanks !
    I have never heard of this 'ceiling function' before
    I found it by myself
    I took D_n = -\left\lceil {\frac{1 - {\sqrt {8n + 1}}}{2}} \right\rceil

    instead of C_n = \left\lceil {\frac{{\sqrt {8n + 1} - 1}}{2}} \right\rceil
    because it is not exactly the same (if I am not mistaken)
    C_n = 1, 1, 2, 2, 2, 3, 3, 3, 3 ...
    D_n = 1,2, 2, 3, 3, 3, ...
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  6. #6
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    Quote Originally Posted by sethu1989 View Post
    Yes, I want to know how you got it.
    Not knowing the 'ceiling function' here is how I got it
    First you have to know that
    \sum_{j=1}^{k}j = \frac{k(k+1)}{2}

    u_n = k for every n integer such as \frac{(k-1)k}{2} < n \leq \frac{k(k+1)}{2}

    n being given, to find k you must have both inequations
    (k-1)k - 2n < 0 and k(k+1) - 2n \geq 0

    k^2 - k - 2n and k^2 + k - 2n are parabolas with the same discriminant 8n+1

    k^2 - k - 2n < 0 for k < \frac{1+\sqrt{8n + 1}}{2}
    k^2 + k - 2n \geq 0 for k \geq \frac{-1+\sqrt{8n + 1}}{2}

    Therefore \frac{-1+\sqrt{8n + 1}}{2} \leq k < \frac{1+\sqrt{8n + 1}}{2}

    There is only one integer k satisfying these inequations because

    \frac{1+\sqrt{8n + 1}}{2} - \frac{-1+\sqrt{8n + 1}}{2} = 1
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