The continued fraction $2 + \cfrac{1}{4+\cfrac{1}{6+\cfrac{1}{6+\cfrac{1}{6+\cdots}}}} $

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Question: I am given the linear form of a simple continued fraction $[2;4,6,6,6,\ldots]$ which I believe may be written in fraction form as $$ q = 2 + \cfrac{1}{4 + \cfrac{1}{6 + \cfrac{1}{6 + \cfrac{1}{6 + \cdots}}}} $$ How can I compute $q$ explicitly? I would like to think that $ q=\frac{a+\sqrt{b}}{c} $ , it's periodic, but I am not certain. I feel it is also fair to ask if there are any online symbolic calculators that can solve similar expression ?


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Let

$$x=\frac{1}{6+\frac{1}{6+\frac{1}{6+...}}}$$

Then

$$\frac{1}{x}=6+\frac{1}{6+\frac{1}{6+\frac{1}{6+...}}}=6+x$$

This works because the continued fraction is infinite, so you can simply substitute the original value of $x$ for the repeated continued fraction. Then multiply both sides by $x$ to obtain

$$1=6x+x^2$$

This is quadratic, so you can solve for x. But since you are multiplying by $x$, there might be an incorrect solution... make sure you check this at the end to ensure the correct solution. Then you can simplify from there for the entire continued fraction by saying

$$y=2+\frac{1}{4+x}$$

and substituting your value for $x$.

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First of all, let's consider $$x=\frac{1}{6+\frac{1}{6+\frac{1}{6+\ldots}}}$$ $$x=\frac{1}{6+x}$$ $$x^2+6x-1=0$$ $$x=\frac{-6\pm\sqrt{6^2-4(1)(-1)}}{2(1)}$$ $$x=-3\pm\sqrt{10}$$ Now, setting $x=-3+\sqrt{10}$, we get $$q=2+\frac{1}{4+\frac{1}{6+\frac{1}{6+\ldots}}}$$ $$q=2+\frac{1}{4+(-3+\sqrt{10})}$$ $$q=2+\frac{1}{\sqrt{10}+1}$$ $$q=2+\frac{\sqrt{10}-1}{(\sqrt{10}+1)(\sqrt{10}-1)}$$ $$q=2+\frac{\sqrt{10}-1}{9}$$ $$q=\frac{17+\sqrt{10}}{9}$$ Setting another value, $x=-3-\sqrt{10}$, we get negative value of $q$ e.i. $q=\frac{-17+\sqrt{10}}{9}$ which is not acceptable

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Note that $$x=6+\frac{1}{6+\frac{1}{6+\frac{1}{6+...}}}$$ satisfies $$x=6+1/x$$ with the acceptable answer $$ x=3+\sqrt {10}$$

Now we substitute into $y= 4+1/x$ to get $y=1+\sqrt {10}$.

Finally we plug in this value for $y$ in $$q=2+1/y$$ to get $$q=\frac {17+\sqrt {10}}{9}$$