Official solution
f(x + y) = f(x)f(y)
\[\sum\limits_{x = 1}^\infty {f\left( x \right)} = 2\]
\[ \Rightarrow \] f(1) + f(2) + f(3) + ........\[\infty \] = 2 ....(1)
On f(x + y) = f(x) f(y)
* Put x = 1, y = 1
f(2) = (f(1))2
* Put x = 2, y = 1
f(3) = f(2). f(1) = f((1))3
* Put x = 2, y = 2
f(4) = f((2))2 = f((1))4
Now put these values in equation (1)
f(1) + f((1))2 + f((1))3 + ....... = 2
\[ \Rightarrow \] \[{{f\left( 1 \right)} \over {1 - f\left( 1 \right)}}\] = 2
\[ \Rightarrow \] f(1) = \[{2 \over 3}\]
Now f(2) = \[{\left( {{2 \over 3}} \right)^2}\]
and f(4) = \[{\left( {{2 \over 3}} \right)^4}\]
\[ \therefore \] \[{{f\left( 4 \right)} \over {f\left( 2 \right)}}\]
= \[{{{{\left( {{2 \over 3}} \right)}^4}} \over {{{\left( {{2 \over 3}} \right)}^2}}}\] = \[{4 \over 9}\]