250 Followers Problem

We will try to solve this problem in parts to make the simplification easy to understand.

Let's work out the numerator first, using method to evaluate an arithmetic-geometric progression as follows: $\begin{aligned} \sqrt{(5\sqrt{5}+5)\sqrt{(5\sqrt{5}+5)^{2}\sqrt{(5\sqrt{5}+5)^{3}\sqrt{\ldots}}}} = & (5\sqrt{5}+5)^{\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\ldots} \\ = & [5(\sqrt{5}+1)]^2 \end{aligned}$

Note :Let's see why $\dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\ldots=2$ . $\begin{aligned} \mathbf{S} = & \dfrac{1}{2}+\dfrac{2}{4}+\dfrac{3}{8}+\ldots \quad \quad \quad \ldots (1) \\ \implies 2\mathbf{S} = & \dfrac{1}{1}+\dfrac{2}{2}+\dfrac{3}{4}+\ldots \quad \quad \quad \ldots (2) \\ \therefore \mathbf{S}= & 1+ \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\ldots \quad \quad \quad [\text{Subtracting} ~ (1) ~ \text{from} ~ (2)] \end{aligned}$ Now this a geometric progression ,with $a=1$ and $r=\dfrac{1}{2}$ ,and thus evaluates to $\dfrac{1}{1-\frac{1}{2}}=2~ .$

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Next,note that $6+2\sqrt5=(\sqrt5+1)^2$

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As a last step,recall the identity $\begin{aligned} \color{#3D99F6}{(a-b)}^3= & a^3-b^3-3ab\color{#3D99F6}{(a-b)} \\ \therefore \color{#3D99F6}{(a-b)}= & \sqrt[3]{a^3-b^3-3ab\color{#3D99F6}{(a-b)}} \\ = & \sqrt[3]{a^3-b^3-3ab\sqrt[3]{a^3-b^3-3ab\color{#3D99F6}{(a-b)}}} \\ = & \sqrt[3]{a^3-b^3-3ab\sqrt[3]{a^3-b^3-3ab\sqrt[3]{a^3-b^3-3ab\color{#3D99F6}{(a-b)}}}} \\ = & \ldots \end{aligned}$ Replacing $a$ with $6$ and $b$ with $1$ ,we get $\begin{aligned} 6-1= & \sqrt[3]{6^3-1-3\cdot6\sqrt[3]{6^3-1-3\cdot6\sqrt[3]{6^3-1-3\cdot6\sqrt[3]{\ldots}}}} \\ 5 =& \sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{\ldots}}}}\end{aligned}$

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$\therefore \log_{\sqrt{5}}{\left[\dfrac{\sqrt{(5\sqrt{5}+5)\sqrt{(5\sqrt{5}+5)^{2}\sqrt{(5\sqrt{5}+5)^{3}\sqrt{\ldots}}}}}{(6+2\sqrt{5})\sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{215-18\sqrt[3]{\ldots}}}}}\right]} \\ =\log_{\sqrt5}{\left[\dfrac{25(\sqrt5+1)^2}{(\sqrt5+1)^2\cdot5}\right]}=\log_{\sqrt5}{5}=\boxed{\boxed{2}}$

Solving for each part, the top part of the fraction is (5sqrt(5)+5)^(1/2+2/4+3/8+....) Let's solve this. Let S equal this sum. Then 2S=1+1+3/4+4/8... Subtracting these, our sum is equivalent to 1+1/2+1/4... Using the formula a/(1-r) for infinite geometric series, 1/(1-1/2)=2, so the top part is 25(6+2sqrt(5)). Next for the bottom. Letting x be the sum, x=cbrt(215-18x). Cubing both sides, we get x^3+18x-215=0. Using the rational root theorem with factors of 215, we get 5 to be a solution. Therefore x=5. Overall we get log(25/5)/log(sqrt(5)), after the 6+2sqrt(5) cancel out and with log base b of a equaling log a/log b. Since log(sqrt(5))=log(5^1/2) and log a^b=b log a, our final answer is 1/(1/2)=2.