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Question-151790




Question Number 151790 by mathdanisur last updated on 23/Aug/21
Answered by Olaf_Thorendsen last updated on 24/Aug/21
u_1  = (√(99))  u_n  = (√(102−3n+u_(n−1) ))  u_(33)  = (√(3+(√(6+(√(9+...+(√(96+(√(99))))))))))  u_(33)  ≈ 2,469925717
$${u}_{\mathrm{1}} \:=\:\sqrt{\mathrm{99}} \\ $$$${u}_{{n}} \:=\:\sqrt{\mathrm{102}−\mathrm{3}{n}+{u}_{{n}−\mathrm{1}} } \\ $$$${u}_{\mathrm{33}} \:=\:\sqrt{\mathrm{3}+\sqrt{\mathrm{6}+\sqrt{\mathrm{9}+…+\sqrt{\mathrm{96}+\sqrt{\mathrm{99}}}}}} \\ $$$${u}_{\mathrm{33}} \:\approx\:\mathrm{2},\mathrm{469925717}\: \\ $$

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