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Arithmetic Sequences

Find a closed form formula for the sequence {1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,…}

    \[\begin{tabular}{c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c} n&\boxed{1}&2&\boxed 3&4&5&\boxed 6&7&8&9&\boxed{10}&11&12&13&14&\boxed{15} \\ \hline a_n&1&2&2&3&3&3&4&4&4&4&5&5&5&5&5 \end{tabular}\]

Looking at the a_n‘s associated with the boxed n‘s, we see that for the boxed numbers, n = \sum_i^{a_n} i. For example

    \[1 = 1, 3 = 1+2, 6 = 1+2+3, 10 = 1+2+3+4, 15 = 1+2+3+4+5\]

This naturally produces lower and upper bounds on the range of n for some particular a_n.

(1)   \begin{align*} \sup n &= \sum_{i=1}^{a_n} i \\ &= \frac{a_n^2 + a_n}{2} \\ 0 &= a_n^2 + a_n - 2\hat n & \mbox{let } \hat n = \sup n \\ a_n &= \frac{-1 \pm \sqrt{8\hat n + 1}}{2} \\ &= \left\lceil \frac{\sqrt{8 n + 1} - 1}{2} \right\rceil \end{align*}

Because (1) monotonically increases, by bounding the n for the a_n, we’ll have successively shown that our expression satisfies the problem, we only need to show that \sup n - \sup(n-1) = a_n

    \begin{align*} \sup n &= \frac{a_n^2 + a_n}{2} \\ \sup n - \sup(n-1) &= \frac{a_n^2 + a_n}{2} - \frac{(a_n-1)^2 + a_n - 1}{2} \\ &= \frac{a_n^2 + a_n}{2} - \frac{a_n^2 - 2a_n + 1 + a_n-1}{2} \\ &= \frac{a_n^2 + a_n}{2} - \frac{a_n^2 - a_n}{2} \\ &= \frac{2a_n}{2} \\ &= a_n \blacksquare \end{align*}

Posted By Lee

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