Question

√6+√6+√6+√6.................infinity

Answer 1

n n+1i.e 6 factors 2×3 i choosed this becausee their difference is 1 then here n+1 is 3

Answer 2

$\mathcal{\tiny{\boxed{\bold{Heya\:mate. \: Solution\: below}}}}$

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★ $\red{\mathfrak{\: Answer \: ↓↓↓\: }}$★

$\bf{\large{To\:Find \: -}}$$\sf{\sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } }$

$\sf{let \: x \: = \sqrt{6 + \sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } }\:\:\: ....eq \:i}$

$\sf{= > x = \sqrt{6 + (\sqrt{6 + \sqrt{6 + \sqrt{6... \infty } } } } } )$

$\sf{ = > x = \sqrt{6 + x} \: \: \: ... \: \: from \: eq \: i}$

$\bf{\red{Squaring\: both \:sides,}}$

$\sf{= > {x}^{2} = 6 + x}$

$\sf{= > {x}^{2} - x - 6 = 0}$

$\sf{= > {x}^{2} - 3x + 2x - 6 = 0}$

$\sf{= > x(x - 3) + 2(x - 3) = 0}$

$\sf{= > (x - 3)(x + 2 ) = 0}$

$\sf{= > x = - 3 \: \: \: \: \: \: or \: \: \: \: \: \: x = 2}$

But, the value can't be negative.

Therefore, x = 3

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$\mathcal{\green{\bold{Thank \: you..}}}$