If f(x + y) = f(x)f(y) and sum_{x=1}^{infty} | vedclass

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(B) Given the functional equation $f(x + y) = f(x)f(y)$.For $x = 1, y = 1$,$f(2) = f(1)^2$.For $x = 2, y = 1$,$f(3) = f(2)f(1) = f(1)^3$.In general,$f

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$\frac{4}{9}$

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(B) Given the functional equation $f(x + y) = f(x)f(y)$.For $x = 1, y = 1$,$f(2) = f(1)^2$.For $x = 2, y = 1$,$f(3) = f(2)f(1) = f(1)^3$.In general,$f(x) = f(1)^x$.Given the infinite geometric series $\sum_{x=1}^{\infty} f(x) = 2$,we have $f(1) + f(1)^2 + f(1)^3 + \dots = 2$.This is a geometric series with first term $a = f(1)$ and common ratio $r = f(1)$.The sum is given by $\frac{a}{1 - r} = 2$,so $\frac{f(1)}{1 - f(1)} = 2$.$f(1) = 2 - 2f(1) \implies 3f(1) = 2 \implies f(1) = \frac{2}{3}$.Now,we need to find $\frac{f(4)}{f(2)}$.Since $f(x) = f(1)^x$,we have $\frac{f(4)}{f(2)} = \frac{f(1)^4}{f(1)^2} = f(1)^2$.Substituting $f(1) = \frac{2}{3}$,we get $\left(\frac{2}{3}\right)^2 = \frac{4}{9}$.

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

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