Key Concepts and Formulas

• Functional Equation: The given equation $5 f(x+y) = f(x) \cdot f(y)$ is a type of functional equation that often leads to exponential functions.
• Properties of Differentiable Functions: Differentiability allows us to use calculus techniques, such as differentiation, to find the form of the function.
• Geometric Progression: If a function generates terms where each term is a constant multiple of the previous one, it forms a geometric progression.

Step-by-Step Solution

Step 1: Analyze the Functional Equation and Determine the Form of $f(x)$. We are given the functional equation $5 f(x+y) = f(x) \cdot f(y)$ for all $x, y \in \mathbb{R}$, and $f: \mathbb{R} \rightarrow (0, \infty)$. Let $y=0$. Then, $5 f(x) = f(x) \cdot f(0)$. Since $f(x) > 0$, we can divide by $f(x)$: $5 = f(0)$. Now, let's differentiate the functional equation with respect to $y$, treating $x$ as a constant: $\frac{d}{dy} [5 f(x+y)] = \frac{d}{dy} [f(x) \cdot f(y)]$ $5 f'(x+y) \cdot \frac{d}{dy}(x+y) = f(x) \cdot f'(y)$ $5 f'(x+y) = f(x) \cdot f'(y)$ Now, let $y=0$: $5 f'(x) = f(x) \cdot f'(0)$ Let $f'(0) = c$. Then, $5 f'(x) = c f(x)$. This is a first-order linear differential equation: $f'(x) - \frac{c}{5} f(x) = 0$. The solution to this differential equation is $f(x) = A e^{\frac{c}{5} x}$, where $A$ is a constant of integration. We know $f(0) = 5$, so $A e^{\frac{c}{5} \cdot 0} = 5$, which means $A = 5$. Thus, $f(x) = 5 e^{\frac{c}{5} x}$. Let $k = \frac{c}{5}$. Then, $f(x) = 5 e^{kx}$. Substitute this form back into the original functional equation to verify: $5 (5 e^{k(x+y)}) = (5 e^{kx}) \cdot (5 e^{ky})$ $25 e^{kx+ky} = 25 e^{kx+ky}$ This confirms our form of $f(x)$.

Step 2: Use the Given Value of $f(3)$ to Find the Value of $k$. We are given that $f(3) = 320$. Using the form $f(x) = 5 e^{kx}$: $f(3) = 5 e^{k \cdot 3} = 320$ $e^{3k} = \frac{320}{5} = 64$ To find $k$, we can take the natural logarithm of both sides: $\ln(e^{3k}) = \ln(64)$ $3k = \ln(64)$ $k = \frac{\ln(64)}{3} = \frac{\ln(4^3)}{3} = \frac{3 \ln(4)}{3} = \ln(4)$ So, the function is $f(x) = 5 e^{x \ln(4)} = 5 e^{\ln(4^x)} = 5 \cdot 4^x$.

Step 3: Calculate the Values of $f(n)$ for $n=0, 1, 2, 3, 4, 5$. We have $f(x) = 5 \cdot 4^x$.

• $f(0) = 5 \cdot 4^0 = 5 \cdot 1 = 5$.
• $f(1) = 5 \cdot 4^1 = 5 \cdot 4 = 20$.
• $f(2) = 5 \cdot 4^2 = 5 \cdot 16 = 80$.
• $f(3) = 5 \cdot 4^3 = 5 \cdot 64 = 320$. (This matches the given information, serving as a check).
• $f(4) = 5 \cdot 4^4 = 5 \cdot 256 = 1280$.
• $f(5) = 5 \cdot 4^5 = 5 \cdot 1024 = 5120$.

Step 4: Calculate the Sum \sum_\limits{n=0}^{5} f(n). We need to find the sum of the terms calculated in Step 3: $\sum_{n=0}^{5} f(n) = f(0) + f(1) + f(2) + f(3) + f(4) + f(5)$ $\sum_{n=0}^{5} f(n) = 5 + 20 + 80 + 320 + 1280 + 5120$ This is a geometric series with first term $a = 5$, common ratio $r = 4$, and $N=6$ terms. The sum of a geometric series is given by $S_N = a \frac{r^N - 1}{r - 1}$. $\sum_{n=0}^{5} f(n) = 5 \cdot \frac{4^6 - 1}{4 - 1}$ $4^6 = (4^3)^2 = 64^2 = 4096$ $\sum_{n=0}^{5} f(n) = 5 \cdot \frac{4096 - 1}{3}$ $\sum_{n=0}^{5} f(n) = 5 \cdot \frac{4095}{3}$ $\frac{4095}{3} = 1365$ $\sum_{n=0}^{5} f(n) = 5 \cdot 1365$ $5 \cdot 1365 = 6825$

Alternatively, by direct summation: $5 + 20 = 25$ $25 + 80 = 105$ $105 + 320 = 425$ $425 + 1280 = 1705$ $1705 + 5120 = 6825$

Common Mistakes & Tips

• Incorrectly solving the functional equation: Ensure you correctly use differentiation or substitution to find the general form of $f(x)$. Mistakes in finding $f(0)$ or $f'(0)$ will lead to an incorrect function.
• Algebraic errors in solving for the constant: When solving for $k$ using $f(3)=320$, be careful with logarithms and exponents. For example, $\ln(64)$ can be simplified to $3 \ln(4)$.
• Miscalculating the sum of the geometric series: Double-check the number of terms ($N=6$ for $n=0$ to $5$) and the formula for the sum of a geometric progression.

Summary

The problem involves a functional equation that, for a differentiable function with a positive range, leads to an exponential form. By using the given condition $f(3)=320$ and $f(0)=5$, we determined the specific function to be $f(x) = 5 \cdot 4^x$. We then calculated the values of $f(n)$ for $n=0$ to $5$ and summed them up. The sum forms a geometric progression, which was calculated using the standard formula for the sum of a geometric series, yielding the final answer.

The final answer is $\boxed{6825}$.