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Math Help - nth term of a sequence

  1. #1
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    nth term of a sequence

    Any suggestions for an approach to proving that the nth term of the following sequence:

    1,2,2,3,3,3,4,4,4,4,5,5,5...

    is [ ((2*n)^(1/2) + (1/2)) ], where [x] denotes the floor value of x and ^ denotes to the power of
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  2. #2
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    Quote Originally Posted by p vs np View Post
    Any suggestions for an approach to proving that the nth term of the following sequence:

    1,2,2,3,3,3,4,4,4,4,5,5,5...

    is [ ((2*n)^(1/2) + (1/2)) ], where [x] denotes the floor value of x and ^ denotes to the power of
    let A_n be the n-th term of the sequence. it's easy to see that your sequence can be written as: A_{\frac{m(m-1)}{2} + k}=m, \ \ m \geq 1, \ 1 \leq k \leq m. now let:

    \frac{m(m-1)}{2}+k=n, where m \geq 1 and 1 \leq k \leq m. so A_n=m, and: 8n=4m^2-4m+8k=(2m-1)^2+8k-1. thus : 8n > (2m-1)^2. \ \ \ (1)

    also we have: 8n=(2m+1)^2-8(m-k)-1. thus: 8n < (2m+1)^2. \ \ \ \ (2)

    taking square root in (1) and (2) will give us: m < \sqrt{2n} + \frac{1}{2} < m+1, which means: \left \lfloor \sqrt{2n} + \frac{1}{2} \right \rfloor=m=A_n.
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  3. #3
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    Here is another way to generate that sequence: a_n  = \left\lceil {\frac{{ - 1 + \sqrt {8n + 1} }}{2}} \right\rceil .
    That is of course using the ceiling function.
    Notice the sequence changes on the triangular numbers, 1,3,6,10,; \frac{{n(n + 1)}}{2}.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Cool

    First of all, I have to say that this is a very good exercise.

    <br />
\begin{array}{cccccccccccc}<br />
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \cdots \\<br />
a_{n} & 1 & 2 & 2 & 3 & 3 & 3 & 4 & 4 & 4 & 4 & \cdots\\<br />
 & * & & * & & & * & & & & * & <br />
\end{array}<br />

    Consider the terms marked with *, for these numbers, we have n=\sum\limits_{i=1}^{a_{n}}i=\frac{a_{n}(a_{n}+1)}  {2}.

    Now, let a_{n_{0}}\leq a_{n}\leq a_{n^{0}}, where n^{0}\geq n is the smallest number which is not less than n and satisfies n^{0}=\frac{a_{n^{0}}(a_{n^{0}}+1)}{2} (in short, lets say * property) and n_{0}\leq n is the greatest number which is less than n and satisfies * property.
    Therefore, solving this parabola with, we have a_{n^{0}}=\frac{-1\pm\sqrt{1+8n^{0}}}{2}. Note that the desired term is the positive one, hence a_{n^{0}}=\frac{-1+\sqrt{1+8n^{0}}}{2}. Thus, the solution is completed when n has * property.

    If n does not have * property, then we see that n_{0}<n<n^{0}, you can easily show that a_{n}=\left\lceil\frac{-1+\sqrt{1+8n}}{2}\right\rceil, because of the definitions of n_{0},n^{0}, being a_{n_{0}}+1=a_{n^{0}}=a_{n} and the sequezing n_{0}=\frac{a_{n_{0}}(a_{n_{0}}+1)}{2}\leq n\leq\frac{a_{n^{0}}(a_{n^{0}}+1)}{2}=n^{0}.

    \therefore Plato's answer is right!
    Last edited by bkarpuz; September 9th 2008 at 10:02 PM. Reason: [math] n=\sum\limits_{i=1}^{a_{n}}i=\frac{a_{n}(a_{n}+1)}{2}. [/math] is corrected.
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