Suppose quad f : R rightarrow(0, infty) be a | vedclass

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Question

$5 f ( x + y )= f ( x ) \cdot f ( y )$

$5 f (0)= f (0)^2 \Rightarrow f (0)=5$

$5 f ( x +1)= f ( x ) \cdot f (1)$

$\frac{ f ( x +1)}{ f ( x )}

Vedclass · Accepted Answer

$6825$

Vedclass · Answer

$5 f ( x + y )= f ( x ) \cdot f ( y )$

$5 f (0)= f (0)^2 \Rightarrow f (0)=5$

$5 f ( x +1)= f ( x ) \cdot f (1)$

$\frac{ f ( x +1)}{ f ( x )}=\frac{ f (1)}{5}$

$\frac{ f (1)}{ f (0)} \cdot \frac{ f (2)}{ f (1)} \cdot \frac{ f (3)}{ f (2)}=\left(\frac{ f (1)}{5}\right)^3$

$\frac{320}{5}=\frac{( f (1))^3}{5^3} \Rightarrow f (1)=20$

$5 f ( x +1)=20 \cdot f ( x ) \Rightarrow f ( x +1)=4 f ( x )$

$\sum \limits_{ n =0}^5 f ( n )=5+5.4+5.4^2+5.4^3+5.4^4+5.4^5$

$=\frac{5\left[4^6-1\right]}{3}=6825$

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

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