Question Number 151790 by mathdanisur last updated on 23/Aug/21
Answered by Olaf_Thorendsen last updated on 24/Aug/21
$${u}_{\mathrm{1}} \:=\:\sqrt{\mathrm{99}} \\ $$$${u}_{{n}} \:=\:\sqrt{\mathrm{102}−\mathrm{3}{n}+{u}_{{n}−\mathrm{1}} } \\ $$$${u}_{\mathrm{33}} \:=\:\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+…+\sqrt{\mathrm{96}+\sqrt{\mathrm{99}}}}}} \\ $$$${u}_{\mathrm{33}} \:\approx\:\mathrm{2},\mathrm{469925717}\: \\ $$