Infinite Nested Radical

We know that $\dfrac 14+\dfrac 34=1$ , so this Nested radical will look like this: $\sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \dots } } } }=\sqrt{1\sqrt{1\sqrt{1\sqrt{1\dots}}}}=\boxed 1$

Let $x = \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \cdots } } } }$

Using recurrent relation, we will get $x = \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 }x}$ , or simply $4x^2-3x-1 = 0$ . Factorizing it gives $(4x+1)(x-1) = 0$ hence $x = \frac{-1}{4}, 1$ , but as $0 < x$ , hence $x = 1$

two solns are possible 1 and 1/4

Let the equation be y.......... since its an infinite series therefor addition or subtraction of some terms would not affect the sum of series. Therefor............. Y=√{1/4 + 3/4Y}............now.... Y^2 = 1/4 + 3/4Y.........now........ 4Y^2 -3Y -1 = 0............ Hence roots of eq. are -2/3 and -1/6 Since sum of positive no cannot be negative Therefor ans = 2/3 ~1.

let $a=\sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \frac { 1 }{ 4 } +\frac { 3 }{ 4 } \sqrt { \dots } } } }$

so $a=\sqrt{\left(\dfrac 14+\dfrac 34\right)a}\\a^2=\left(\dfrac 14+\dfrac 34\right)\\4\times a^2-3a-1=0\quad \implies(4a+1)(a-1)=0$

and $a>0$ , so $\boxed{a=1}$

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[Edits - Moderator]