√2012+√2012+√2012+⋯
This is not an infinite series but is limited to the 2012th term. How can this expression be simplified?
Answer
xn=√2012+xn−1⟺x2n=2012+xn−1⟺x2n−2012=xn−1
And setting x0=0 yields the relation x21−2012=0∴(x22−2012)2−2012=0∴((x23−2012)2−2012)2−2012=0 which you can keep unravelling until you get a polynomial equation in x2012, 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→∞, un becomes the same as un−1 so you then get the equation x2∞−x∞−2012=0 which is a quadratic equation you can solve to get x∞=12(1+√8049) and it should be the case that x2012≈x∞≈45.
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