1.

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Answer:

3

Step-by-step EXPLANATION:

Let

y = \sqrt{6 + \sqrt{6 + \sqrt{6 + ...... + \infty } } }

= > y = \sqrt{6 + y}

= > {y}^{2} = 6 + y

= > {y}^{2} - y - 6 = 0

= > {y}^{2} - 3y + 2y - 6 = 0

= > y(y - 3) + 2(y - 3) = 0

= > (y - 3)(y + 2) = 0

either \: \: y = 3 \: \: or \: \: y = - 2

since, SQUARE ROOT is a POSITIVE QUANTITY,

so we have y=3


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