# 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.

• Hint: Is there a pattern to the fraction? – Dude156 Feb 14 at 2:08

## 2 Answers

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.

HINT:

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