Question

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

• A. 6875
• B. 6575
• C. 6825
• D. 6528

Answer 1

The Correct Option is C

The functional equation is: 

\[ 5f(x + y) = f(x) \cdot f(y) \]

Substitute \(y = 0\):

\[ 5f(x + 0) = f(x) \cdot f(0) \quad \Rightarrow \quad 5f(x) = f(x) \cdot f(0) \]

Divide by \( f(x) \) (since \( f(x) > 0 \)):

\[ f(0) = 5 \]

Substitute \(y = 1\):

\[ 5f(x + 1) = f(x) \cdot f(1) \]

Divide by \( f(x) \):

\[ \frac{f(x + 1)}{f(x)} = f(1) \]

This shows \( f(x + 1) = f(x) \cdot c \), where \( c = f(1) \).

Using the recursive relation, we get:

\[ f(n) = f(0) \cdot c^n = 5 \cdot c^n \]

From \( f(3) = 320 \):

\[ f(3) = f(0) \cdot c^3 = 5 \cdot c^3 \]

\[ 320 = 5 \cdot c^3 \quad \Rightarrow \quad c^3 = 64 \quad \Rightarrow \quad c = 4 \]

Thus, \( f(1) = 5 \cdot c = 5 \cdot 4 = 20 \).

Now compute:

\[ \sum_{n=0}^{5} f(n) \]

\[ f(n) = 5 \cdot 4^n \]

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

The summation inside the parentheses is a geometric series:

\[ \text{Sum} = \frac{4^6 - 1}{4 - 1} = \frac{4096 - 1}{3} = \frac{4095}{3} = 1365 \]

\[ \sum_{n=0}^{5} f(n) = 5 \cdot 1365 = 6825 \]

Conclusion: The value of \( \sum_{n=0}^{5} f(n) \) is 6825. Therefore, the correct answer is \( \boxed{6825} \).