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$$ \sqrt{2012+\sqrt{2012+\sqrt{2012+\cdots}}} $$

This is not an infinite series but is limited to the 2012th term. How can this expression be simplified?

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    $\begingroup$ Why do you think it can be? You can compute an approximate value. It will be very close to the limit, which is $\frac 12(1+\sqrt{8049})$ $\endgroup$ – Ross Millikan Jul 9 '15 at 15:45
  • $\begingroup$ I checked using Python and Ross Milikan is right. It does converge if infinite $\endgroup$ – man and laptop Jul 9 '15 at 15:51
  • $\begingroup$ The sum was given by one of my teacher. We are not allowed to use calculator or any other device. $\endgroup$ – Dhiraj Barnwal Jul 9 '15 at 15:57
  • $\begingroup$ what do you mean by simplified? $\endgroup$ – man and laptop Jul 9 '15 at 16:01
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    $\begingroup$ If we define a sequence $(u_n)$, we have $u_1=\sqrt{2012}$ and $u_{n+1}^2=2012 + u_n$. We need to compute $u_{2012}$... Not sure it helps a lot to find a closed form. $\endgroup$ – mathcounterexamples.net Jul 9 '15 at 16:03
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$$x_n = \sqrt{2012 + x_{n-1}} \iff x_n^2 = 2012 + x_{n-1} \iff x_n^2 - 2012 = x_{n-1}$$ And setting $$x_{0} = 0 $$ yields the relation $$x_1^2 - 2012 = 0 \therefore (x_2^2 - 2012)^2 - 2012 = 0 \therefore ((x_3^2 - 2012)^2 - 2012)^2 - 2012 = 0$$ which you can keep unravelling until you get a polynomial equation in $x_{2012}$, and maybe a CAS can then express a nice closed form solution for the roots of that polynomial (which is somehow simpler than what you wrote). Otherwise you can solve it approximately as Ross Milikan did.

The way his solution works is that you notice that as $n \rightarrow \infty$, $u_n$ becomes the same as $u_{n-1}$ so you then get the equation $$x_\infty^2 - x_\infty - 2012 = 0$$ which is a quadratic equation you can solve to get $$x_\infty = {1 \over 2}(1 + \sqrt{8049}) $$ and it should be the case that $x_{2012} \approx x_\infty \approx 45$.

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