If $$x = 1 + \frac{1}{2^2} + \frac{1}{2^3} + ....\infty$$ and  $$y = x + \frac{1}{2} + \frac{x}{9} + \frac{1}{18} + \frac{x}{81} + \frac{1}{162} + ....\infty$$, then

for an infinite series , SUM$$=\ \frac{\ a}{\left(1-r\right)}.$$

So, $$y=\ \frac{\ x}{\left(1-\ \frac{\ 1}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(1-\ \frac{\ 1}{9}\right)}.$$

or,$$y=\ \frac{\ x}{\left(\frac{\ 8}{9}\right)}+\ \frac{\ \ \frac{\ 1}{2}}{\left(\frac{\ 8}{9}\right)}.$$

or,$$y=\ \frac{9\ x}{8}+\frac{\ \ 9}{16}.$$

Again,

$$x=\ \frac{\ \ \frac{\ 1}{2^2}}{\left(1-\ \frac{\ 1}{2}\right)}+1.$$

or,$$x=\frac{\ 3}{2}.$$

So,$$y=\frac{\ 27}{16}+\frac{\ 9}{16}=\frac{\ 36}{16}=\frac{\ 9}{4}=x^2.$$

C is correct choice.