Suppose f: R → (0, ∞) be a differentiable function such that 5f(x +...

Given, equation is:

$5f\left(x + y\right) = f\left(x\right) \cdot f\left(y\right) . . . \left(1\right)$

At $x = 0, y = 0$, we get

$5f\left(0\right) = f{\left(0\right)}^{2} ⇒ f\left(0\right) = 5$

Put $y = 1$ in \left(1\right)$, then we get</p><p>$5f\left(x + 1\right) = f\left(x\right) \cdot f\left(1\right)$</p><p>$⇒ \dfrac{f\left(x + 1\right)}{f\left(x\right)} = \dfrac{f\left(1\right)}{5}$</p><p>$⇒ \dfrac{f\left(1\right)}{f\left(0\right)} \cdot \dfrac{f\left(2\right)}{f\left(1\right)} \cdot \dfrac{f\left(3\right)}{f\left(2\right)} = {\left(\dfrac{f\left(1\right)}{5}\right)}^{3}$</p><p>$⇒ \dfrac{320}{5} = \dfrac{{\left(f\left(1\right)\right)}^{3}}{{5}^{3}} ⇒ f\left(1\right) = 20$</p><p>$∴ 5f\left(x + 1\right) = 20 \cdot f\left(x\right)$</p><p>$⇒ f\left(x + 1\right) = 4f\left(x\right)$</p><p>So,</p><p>$f\left(1\right) = 4f\left(0\right) = 4 \cdot 5f\left(2\right) = 4f\left(1\right) = {4}^{2} \cdot 5f\left(3\right) = 4f\left(2\right) = {4}^{3} \cdot 5f\left(4\right) = 4f\left(3\right) = {4}^{4} \cdot 5f\left(5\right) = 4f\left(4\right) = {4}^{5} \cdot 5$</p><p>$\mathop{∑}_{n = 0}^{5}f\left(n\right) = 5 + 5 \cdot 4 + 5 \cdot {4}^{2} + 5 \cdot {4}^{3} + 5 \cdot {4}^{4} + 5 \cdot {4}^{5}$</p><p>$⇒ \mathop{∑}_{n = 0}^{5}f\left(n\right) = \dfrac{5\left({4}^{6} - 1\right)}{3} = 6825$