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:
6875
6575
6825
6528
The Correct Option is C
Solution and Explanation
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 \( \boxed{6825} \).
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