\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ 1+\dfrac{1}{x}=1+\dfrac{1}{1+\frac{1}{2+...}}\\ \)

First we let x = {1;2,1,2,1,...}, then 1+1/x = {1;1,2,1,2,1,2,...}

Then we find the value of x.

\(x=1+\dfrac{1}{2+\frac{1}{1+...}}\\ x=1+\dfrac{1}{2+x}\\ (x-1)(2+x)=1\\ x^2+x-3=0\\ (x+\dfrac{1}{2})^2=\dfrac{13}{4}\\ x+\dfrac{1}{2}=\pm\dfrac{\sqrt{13}}{2}\\ x=\dfrac{-1+\sqrt{13}}{2}\text{ or }x=\dfrac{-1-\sqrt{13}}{2}(rej.)\\ \)

The value of the original expression is 1+1/x which is:

\(1+\dfrac{2}{\sqrt{13}-1}\\ =\dfrac{\sqrt{13}+1}{\sqrt{13}-1}\\ =\dfrac{14+2\sqrt{13}}{12}\\ =\dfrac{7+\sqrt{13}}{6}\)

:D

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