If $$f(x+y) = f(x)f(y)$$ and $$\sum_{x=1}^{\infty} f(x) = 2$$, $$x, y \in N$$, where $$N$$ is the set of all natural numbers, then the value of $$\frac{f(4)}{f(2)}$$ is:

We are given the functional equation $$f(x+y)=f(x)\,f(y)$$ for all natural numbers $$x, y$$.

First, put $$y=1$$. We obtain $$f(x+1)=f(x)\,f(1)$$. Let us denote $$c=f(1)$$ for convenience.

Now we prove by induction that $$f(n)=c^{\,n}$$ for every natural number $$n$$.

Base case: for $$n=1$$ we have $$f(1)=c=c^{\,1}$$, which is true.

Induction step: assume $$f(k)=c^{\,k}$$ for some $$k\in\mathbb N$$. Then

$$f(k+1)=f(k)\,f(1)=c^{\,k}\,c=c^{\,k+1},$$

so the formula holds for $$k+1$$ as well. Hence by mathematical induction $$f(n)=c^{\,n}$$ for all $$n\in\mathbb N$$.

Next, the question supplies the infinite series condition

$$\sum_{x=1}^{\infty} f(x)=2.$$

Substituting $$f(x)=c^{\,x}$$, this series becomes

$$\sum_{x=1}^{\infty} c^{\,x}=2.$$

We recall the standard formula for the sum of an infinite geometric progression with first term $$a$$ and common ratio $$r$$, where $$|r|<1$$:

$$\sum_{n=0}^{\infty} a\,r^{\,n}=\frac{a}{1-r}.$$

In our sum the first term is $$c$$ and the common ratio is also $$c$$, so

$$\sum_{x=1}^{\infty} c^{\,x}=\frac{c}{1-c}=2.$$

We now solve this simple linear equation for $$c$$:

$$\frac{c}{1-c}=2 \;\;\Longrightarrow\;\; c = 2(1-c)$$

$$\Longrightarrow\;\; c = 2 - 2c$$

$$\Longrightarrow\;\; 3c = 2$$

$$\Longrightarrow\;\; c = \frac{2}{3}.$$

Because $$c=\frac23$$ satisfies $$|c|<1$$, the derivation is consistent. Therefore

$$f(n)=\left(\frac{2}{3}\right)^{\!n}.$$

We can now compute the required ratio. First,

$$f(2)=\left(\frac{2}{3}\right)^{\!2}=\frac{4}{9},$$

and

$$f(4)=\left(\frac{2}{3}\right)^{\!4}=\frac{16}{81}.$$

Hence

$$\frac{f(4)}{f(2)}=\frac{\dfrac{16}{81}}{\dfrac{4}{9}}=\frac{16}{81}\times\frac{9}{4}=\frac{4}{9}.$$

Hence, the correct answer is Option 4.