$y=\frac{1}{2+ \frac{1}{3+ \frac{1}{2+ \frac{1}{3+ \dots } } } }$
$y=\frac{1}{2+ \frac{1}{3+y} }$
$y=\frac{1}{ \frac{2y+7}{3+y} }$
$2y^{2}$ + 7y = 3 + y
$2y^{2}$ + 6y – 3 = 0
=$\frac{-6± \sqrt{36-4(-3) 2}} {4}$
=$ \frac {±\sqrt{15}-3} {2}$
y=$\frac {\sqrt{15}-3} {2}$
Hence, Option(4) $\frac {\sqrt{15}-3} {2}$ is the correct choice.