Question 16

Let $$ y = \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+…}}}}$$. Then y equals?


$$ y = \frac{1}{2+\frac{1}{3+\frac{1}{2+\frac{1}{3+…}}}}$$

which is equal to $$ y = \frac{1}{2+\frac{1}{3+y}} $$ solving we get 

$$ y = \frac{3+y}{7+2y} $$ we get

$$2y^2+6y-3=0$$ .

Solution of this equation is $$\frac{\sqrt{15}-3}{2}$$. 

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