Question

Find the value of c%7B1%7D%7B6%2B%5Ccdots%7D%7D%7D%7D%24." id="TexFormula1" title="$6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}}$." alt="$6+\frac{1}{2+\frac{1}{6+\frac{1}{2+\frac{1}{6+\cdots}}}}$." align="absmiddle" class="latex-formula"> Your answer will be of the form $a+b\sqrt{c}$ where no factor of $c$ (other than $1$) is a square. Find $a+b+c$.

Answer 1

Let

$x = 6 + \dfrac1{2 + \dfrac1{6 + \dfrac1{2 + \cdots}}}$

Then

$x = 6 + \dfrac1{2 + \dfrac1x}$

and solving for x gives

$x = 6 + \dfrac x{2x + 1}$

$x = \dfrac{13x + 6}{2x + 1}$

$x(2x+1) = 13x + 6$

$2x^2 + x = 13x + 6$

$2x^2 - 12x - 6 = 0$

$x^2 - 6x - 3 = 0$

By the quadratic formula,

$x = 3 \pm 2\sqrt{3}$

but x must be a positive number, so only the positive square root solution is valid.

$x = 3 + 2\sqrt{3}$

We identify a = c = 3 and b = 2, so a + b + c = 8.