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

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.
