Let $$x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}};$$ then the value of $(2x-1)^2$ equals...

I don't how to start this question. Please help.

  • $\begingroup$ Hint: Is there a pattern to the fraction? $\endgroup$
    – Dude156
    Commented Feb 14, 2019 at 2:08

2 Answers 2


If you only want to know the value that the continued fraction converges to, you use a simple technique:

$$ x=1+\frac{1}{2+\frac{1}{1+\frac{1}{2+\frac{1}{...}}}}=1+\frac{1}{2+\frac{1}{x}} $$

With some manipulation you could come up with the value of $x$, but you want $(2x-1)^2:$

$$ x\left(2+\frac{1}{x} \right)=\left(2+\frac{1}{x} \right)+1 $$

$$ 2x+1=3+\frac{1}{x} $$

Multiply everything by $x$, since we know it's not zero:

$$ 2x^2-2x-1=0 $$

Complete the square that we want by multiplying by $2$: $$ 4x^2-4x-2=4x^2-4x+1-3=(2x-1)^2-3=0 $$

Hence the answer is 3.



Write the expression as $$x=1+\frac{1}{2+\frac{1}{x}}.$$


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