Iterated Roots with Powers

Algebra Level 4

$\sqrt{ 2^1+ \sqrt{ 2^2+ \sqrt{ 2^4 + \sqrt{2^8 + \cdots } } } }$

The value of above expression is in the form $\phi \sqrt{A}$ , where $A$ is an integer. Find $A$ .

Notation : The symbol $\phi$ denotes the golden ratio, $\phi = \dfrac{1+\sqrt5}2$ .

The answer is 2.

1 solution

Rishabh Jain
Feb 15, 2016

Let $\mathcal{S}$ denote the given expression. $\large\mathcal{S}=\sqrt{ 2+ \sqrt{ 2^2+ \sqrt{ 2^4 + \sqrt{2^8 + \ldots } } } }$ Let's multiply both sides by $\color{#D61F06}{\sqrt2}$ $\color{#D61F06}{\sqrt2}\times\mathcal{S}=\sqrt{\color{#D61F06}{2}\times2+ \sqrt{\color{#D61F06}{2^2}\times2^2+ \sqrt{ \color{#D61F06}{2^4}\times2^4+ \sqrt{\color{#D61F06}{2^8}\times2^8 + \ldots } } } }$ $=\sqrt{ 2^2+ \sqrt{ 2^4+ \sqrt{ 2^8 + \sqrt{2^{16} + \ldots } } } }=\mathcal{S}^2-2$ $\Large\Rightarrow \mathcal{S}^2-\sqrt2 \mathcal{S} -2=0$ $\Large \Rightarrow \mathcal{S}=\dfrac{\sqrt2+\sqrt{10}}{2}=\sqrt2\times\dfrac{1+\sqrt{5}}{2}=\sqrt2\times \phi$ $\huge A=\boxed{\color{#007fff}{2}}$

Did it the same way (+1)!

Harsh Khatri - 5 years, 4 months ago

Are you on slack ?

Akshat Sharda - 5 years, 3 months ago

Nope.. !! ......... ..

Rishabh Jain - 5 years, 3 months ago

Wait , when u multiply by √2 , then in √√ it will become 1/4

Aman Rckstar - 5 years, 4 months ago

Can you explain what you mean by "it will become 1/4"?

Calvin Lin Staff - 5 years, 3 months ago

As √2 will multiplied then in root √(2^2) , in second root to will become 2^(1/4)

Aman Rckstar - 5 years, 3 months ago

@Aman Rckstar – Since it will be multiplied in to be $\sqrt{ 2^2 }$ , how did we go from $2^2 \times 2^2$ to 2^(1/4)?

Calvin Lin Staff - 5 years, 3 months ago

Iterated Roots with Powers

The answer is 2.

1 solution

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