JEE MAIN - Mathematics (2023 - 30th January Morning Shift - No. 6)
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 :
6875
6525
6575
6825
Explanation
$$5f(x + y) = f(x).f(y)$$
$$5f(3) = f(1).f(2)$$
$$5f(2) = {(f(1))^2}$$
$$f(10) = 5$$
$$f(1) = 20$$
$$ \Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600$$
$$\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120} $$
$$ = 1750 + 5120 = 6825$$
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