(√(x+(√(4x+(√(16x+(√(64x+...+(√(4^(2019) x+3))))))))))−(√x)=1 x=?


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Question Number 128850 by ruwedkabeh last updated on 10/Jan/21

$\sqrt{x + \sqrt{4 x + \sqrt{16 x + \sqrt{64 x + . . . + \sqrt{4^{2019} x + 3}}}}} - \sqrt{x} = 1$ $x = ?$
	Answered by mindispower last updated on 10/Jan/21

	$\Rightarrow \sqrt{4 x + . . . . . . + \sqrt{4^{2009} x + 3}} = 1 + 2 \sqrt{x}$ $\Rightarrow 4 x + \sqrt{16 x + . . . . . . + \sqrt{4^{2009} x + 3}} = 1 + 4 \sqrt{x} + 4 x = 1 + 2^{2} \sqrt{x} + 4 x$ $\sqrt{16 x + . . . . \sqrt{4^{2009} x + 3}} = 1 + 4 \sqrt{x} .$ $. . t r y t o f i n d l a s t r e l a t i o n u s i n g 2^{n}, 4^{m}$
	Commented by ruwedkabeh last updated on 11/Jan/21

	$t h a n k y o u . i g o t \frac{1}{4^{2019}}$
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