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

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Accepted Answer

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$5f(x + y) = f(x).f(y)$

$5f(3) = f(1).f(2)$

$5f(2) = {(f(1))^2}$

$f(10) = 5$

$f(1) = 20$

$\Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600$

$\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120}$

$= 1750 + 5120 = 6825$

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

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

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