Let f(x) be a function such that f(x + y) = f(x) · f(y) for all x,...

Given,

$f$\left(x + y\right) = f$\left(x\right) \cdot f\left(y\right)$ and $f\left(1\right) = 3$

Now taking $x = 1 \text{ and } y = 1$ we get,

$f\left(1 + 1\right) = f\left(1\right) \cdot f\left(1\right) ⇒ f\left(2\right) = {3}^{2}$

Similarly $f\left(2 + 1\right) = f\left(2\right) \cdot f\left(1\right) ⇒ f\left(3\right) = {3}^{2} \times 3 = {3}^{3}$

And so on $f\left(n\right) = {3}^{n}$

So, $\mathop{∑}_{r = 1}^{n}f\left(r\right) = 3279$

$⇒ f\left(1\right) + f\left(2\right) + f\left(3\right) . . . . . . . f\left(n\right) = 3279$

$⇒ 3 + {3}^{2} + {3}^{3} . . . . . . . + {3}^{n} = 3279$

$⇒ 3 \times \dfrac{{3}^{n} - 1}{3 - 1} = 3279$

$⇒ \dfrac{{3}^{n} - 1}{2} = 1093$

$⇒ {3}^{n} - 1 = 2186$

$⇒ {3}^{n} = 2187$

$⇒ {3}^{n} = {3}^{7}$

$⇒ n = 7$