Suppose $$f: R \to (0, \infty)$$ be a differentiable function such that $$5f(x+y) = f(x) \cdot f(y), \forall x, y \in R$$. If $$f(3) = 320$$, then $$\sum_{n=0}^{5} f(n)$$ is equal to:

Given $$5f(x+y) = f(x) \cdot f(y)$$ for all $$x, y \in \mathbb{R}$$, $$f: \mathbb{R} \to (0, \infty)$$, and $$f(3) = 320$$.

To begin,

Setting $$x = y = 0$$: $$5f(0) = f(0)^2$$

Since $$f(0) > 0$$: $$f(0) = 5$$

Next,

Let $$g(x) = \frac{f(x)}{5}$$. Then:

$$5 \cdot 5g(x+y) = 5g(x) \cdot 5g(y)$$

$$25g(x+y) = 25g(x)g(y)$$

$$g(x+y) = g(x)g(y)$$

Since $$f$$ is differentiable, $$g(x) = a^x$$ for some $$a > 0$$.

So $$f(x) = 5 \cdot a^x$$.

From this,

$$5 \cdot a^3 = 320 \implies a^3 = 64 \implies a = 4$$

$$f(x) = 5 \cdot 4^x$$

Continuing,

$$\sum_{n=0}^{5} f(n) = 5(4^0 + 4^1 + 4^2 + 4^3 + 4^4 + 4^5)$$

$$= 5(1 + 4 + 16 + 64 + 256 + 1024)$$

$$= 5 \times 1365 = 6825$$

The correct answer is Option 3: $$6825$$.