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Sequences in Sequences

Define the sequence (a_n)_n recursively by a_1 = 1 and

     $$ a_{n+1} = \frac{1 + 4a_n + \sqrt{1+24a_n}}{16} ,\hspace{4mm} \mbox{for $n \ge 1$.} $$

Find an explicit formula for a_n in terms of n.

Since this recurrence isn’t linear in a_n, we define a subsequence to get rid of that square root in the hopes that the resulting sequence can be expressed as a linear recurrence.

    \begin{align*} \mbox{Let }b_n &= \sqrt{1+24a_n} \\ b_{n+1}^2 &= 1+24a_{n+1} \\ &= 1 + 24\frac{1 + 4a_n + \sqrt{1+24a_n}}{16} \\ \intertext{Notice that $a_n = \frac{b_n^2-1}{24}$, we can now write $a_n$ in terms of $b_n^2$} &= 1 + 24\frac{1 + 4\frac{b_n^2-1}{24} + b_n}{16} \\ &= 1 + \frac{3}{2} \left( 1 + \frac{b_n^2-1}{6} + b_n \right) \\ &= 1 + \frac{3}{2} + \frac{b_n^2}{4} - \frac{1}{4} + \frac{3b_n}{2} \\ &= \frac{4+6-1+b_n^2 +6b_n}{4} \\ &= \left(\frac{b_n+3}{2}\right)^2 \\ b_n &= \frac{b_n+3}{2} \end{align*}

We now construct a closed form expression for b_n

    \begin{align*} b_1 &= 5 \\ b_2 &= \frac{b_1}{2} + \frac{3}{2} \\ b_3 &= \frac{b_1}{4} + \frac{3}{4} + \frac{3}{2} \\ & \vdots \\ b_k &= b_1 2^{1-k} + 3 \sum_{i=1}^{k-1}2^{-i} \\ &= b_1 2^{1-k} + 3\left(1 - 2^{1-k}\right) \\ &= 3 + \frac{b_1 - 3}{2^{k-1}} \\ &= 3 + 2^{-k} \end{align*}

Therefore, we see by construction that a_n = \frac{\left(3 + 2^{-k}\right)-1}{25}. \blacksquare

Posted By Lee

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