Question

$\huge\purple{\underbrace{\overbrace{\ulcorner{\mid{\overline{\underline{\bold{Explanation\:Required}}}}}}}}$


$\huge{\fbox{\fbox{\orange{\mathfrak{Question \: Answer}}}}}$​

Answer 1

Step-by-step explanation:

y = 1/2+(1/(3+1/(2+1...)

=> y = 1/(2+1/(3+y) )

=> y = (y+3)/(6+2y+1)

=> 6y + 2y² + y = y+3

=> 2y² +6y-3=0

=> y =[ -6+√(6²-4(2)(-3)) ]/2(2)

=> y = [-6 + √(36+24) ]/4

=> y =(-6+√60)/4

=> y = (-6+2√15)/4

=> y = (-3+√15)/2

or y =(-3-√15)/2

Answer 2

Given,

$\displaystyle\longrightarrow\sf{y=\dfrac{1}{2+\dfrac{1}{3+\dfrac{1}{2+\dfrac{1}{3+\ldots}}}}}$

Or,

$\displaystyle\longrightarrow\sf{y=\dfrac{1}{2+\dfrac{1}{3+\left(\dfrac{1}{2+\dfrac{1}{3+\ldots}}\right)}}}$

The term in the bracket is also equal to $\sf{y.}$ Thus,

$\displaystyle\longrightarrow\sf{y=\dfrac{1}{2+\dfrac{1}{3+y}}}$

$\displaystyle\longrightarrow\sf{y=\dfrac{1}{\left(\dfrac{2(3+y)+1}{3+y}\right)}}$

$\displaystyle\longrightarrow\sf{y=\dfrac{3+y}{6+2y+1}}$

$\displaystyle\longrightarrow\sf{y=\dfrac{y+3}{2y+7}}$

$\displaystyle\longrightarrow\sf{2y^2+6y-3=0}$

Then,

$\displaystyle\longrightarrow\sf{y=\dfrac{-6\pm\sqrt{6^2-4\times2\times-3}}{2\times2}}$

$\displaystyle\longrightarrow\sf{y=\dfrac{-6\pm\sqrt{60}}{4}}$

$\displaystyle\longrightarrow\sf{\underline{\underline{y=-\dfrac{3\pm\sqrt{15}}{2}}}}$

$\longrightarrow\sf{\underline{\underline{y=0.44}}\quad\ OR\quad\ \underline{\underline{y=-3.44}}}$