Find the value of

Let's call this fraction x. Now forget about the\(\frac{1}{1+\frac{1}{2+}}\)

Now the rest is exactly the same as x.

Thus, \(\frac{1}{1+\frac{1}{2+x}}=x\)

Doing cross multiplication, we get \(1=x+\frac{x}{2+x}=\frac{x^2+x}{x+2}\)

Doing cross multiplication again(on the left and right side), we get: \(x+2=x^2+x \rightarrow 0=(+1)x^2+(+0)x+(-2).\)

In this case, a=+1, b=0, c=-2. Using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

We get x=\(\frac{+\sqrt{8}}{2}\)or \(\frac{-\sqrt{8}}{2}\)

Since all of 1,1,2,1,1,1,2,1,1,1,2,1.... are all positive, x must be positive.

So, x=\(\frac{\sqrt{8}}{2}=\sqrt{2}\)(approximation: 1.41421356...)

SEE THE ANSWER IN THE MIDDLE OF THE PAGE POSTED UNDER THE NAME "higgsb".

Guest #2:

This is a simple "continued fraction" and has nothing to do with "x" and the "quadratic formula"!!!!!!!; It  simply = 8 /11.

It equals sqrt(2).

Guest #2 has the right idea, but has made a mistake in his cross-multiply stage.

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