Solution

Find the exact value of $[3;\overline{2,6}]$


Suppose $x$ represents the continued fraction in question.

$$x = 3 + \frac{1}{2 + \displaystyle{\frac{1}{\displaystyle{6 + \frac{1}{\displaystyle{2 + \frac{1}{6 + \cdots}}}}}}}$$

Note, if we let $y$ be the fractional part of $x$ (i.e., everything but the $3$), we can express $x$ in two ways:

$$x = 3 + y \quad \quad \textrm{and} \quad \quad x = 3 + \frac{1}{2 + \displaystyle{\frac{1}{\displaystyle{6 + y}}}}$$

Setting these equal to one another and collapsing/simplifying the fractional expression, we discover

$$y = \frac{6+y}{13+2y}$$

Multiplying both sides by $(13+2y)$ yields a quadratic equation,

$$13y+2y^2 = 6 + y$$

which we may readily solve for $y$:

$$2y^2 + 12y - 6 = 0 \quad \quad \longrightarrow \quad \quad y = -3 \pm 2\sqrt{3}$$

Discarding $y = -3-2\sqrt{3}$, since we know $y$ must be positive, we are left with

$$y = -3 + 2\sqrt{3}$$

Finally, remembering that $x=3+y$, we have

$$[3;\overline{2,6}] = 3 + (-3 + 2\sqrt{3}) = 2\sqrt{3}$$