Question

 $z_{1}= \sqrt{3}$
$z_{n+1} = \sqrt{3+ z_{n} }$
sa se afle al 2014-lea termen prin recurenta?

Answer 1

Faci n = 1 => $z_{2} = \sqrt{3+ \sqrt{3} }$;
Faci n = 2 => $z_{3} = \sqrt{3+ \sqrt{3+ \sqrt{3} } } ;$
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Pana ajungi la n = 2013.