Get To The Root Of The Matter

$\begin{array}{l} \sqrt {2\sqrt {4\sqrt {8\sqrt {16\sqrt {32\sqrt {...} } } } } } = {2^{\frac{1}{2}}} \times {2^{\frac{2}{4}}} \times {2^{\frac{3}{8}}} \times {2^{\frac{4}{{16}}}} \times {2^{\frac{5}{{32}}}}.... = {2^{\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{{16}} + \frac{5}{{32}} + ...}} = {2^x}\\ x = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{{16}} + \frac{5}{{32}} + ...(1)\\ \frac{x}{2} = \;\quad \frac{1}{4} + \frac{2}{8} + \frac{3}{{16}} + \frac{4}{{32}} + ...(2)\\ {\rm{subtracting}}\;(1)\;by\;(2),we\;get\\ \frac{x}{2} = \;\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{{16}} + \frac{1}{{32}} + ...\\ \frac{x}{2} = \;\frac{1}{2} + \frac{{\frac{1}{4}}}{{1 - \frac{1}{2}}} = 1\\ x = 2\\ {\rm{Thus,}}\;{\rm{we}}\;{\rm{find}}\;{\rm{the}}\;{\rm{answer}}\;{\rm{4}} \end{array}$

First, note that when we multiply the nested radical by $\sqrt{2\sqrt{2\sqrt{\cdots}}}$ we obtain the nested radical $\sqrt{4\sqrt{8\sqrt{16\sqrt{\cdots}}}}$ .

We can find the value of $\sqrt{2\sqrt{2\sqrt{\cdots}}}$ by setting $x=\sqrt{2\sqrt{2\sqrt{\cdots}}}$ and setting up a recursive equation: $\begin{aligned}x&= \sqrt{2x}\\ x^2&= 2x\\ x&=2\end{aligned}$

We then can find the value of the nested radical in the problem by setting up another recursive equation if $y=\sqrt{2\sqrt{4\sqrt{8\sqrt{\cdots}}}}$

$\begin{aligned}y&=\sqrt{2xy}\\y^2&=2xy\\ y^2&= 4y \\ y&=4\end{aligned}$ so our desired answer is $\boxed{4}$ .

let the expression be equal to x so......sqrt(2^x) = x......squaring we get x^2 = 2^x the only number that fits the equation is 2 or a 4 .

Define the value of the expression recursively. Working from right to left, let $a_1 = 2^k$ , where $k$ is an unspecified large, positive integer. Then $a_2 = \frac{2^k}{2} \sqrt{2^k} = \frac{a_1}{2} \sqrt{a_1}$ , and so on.

Assume the limit of this recursive sequence exists and is finite. As $k$ tends to infinity, $a_{n+1} = a_{n}$ . Therefore the limiting value of $a_n$ is given by $a_n = \frac{a_n}{2} \sqrt{a_n}$ , or $1 = \frac{1}{2} \sqrt{a_n}$ . This means that $\sqrt{a_n} = 2$ , and since the limit is clearly positive, squaring both sides gives $a_n =4$ . In other words, the value of the given expression is $\boxed{4}$ .