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x-4x-16x-4-2019-x-3-x-1-




Question Number 72339 by naka3546 last updated on 27/Oct/19
(√(x + (√(4x + (√(16x + ... + (√(4^(2019) x + 3))))))))  =  (√x) + 1
$$\sqrt{{x}\:+\:\sqrt{\mathrm{4}{x}\:+\:\sqrt{\mathrm{16}{x}\:+\:…\:+\:\sqrt{\mathrm{4}^{\mathrm{2019}} {x}\:+\:\mathrm{3}}}}}\:\:=\:\:\sqrt{{x}}\:+\:\mathrm{1} \\ $$
Commented by naka3546 last updated on 27/Oct/19
x  =  ?
$${x}\:\:=\:\:? \\ $$

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