Thank you for reporting, we will resolve it shortly
Q.
Suppose $f: R \rightarrow(0, \infty)$ 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 $\displaystyle\sum_{n=0}^5 f( n )$ is equal to :
JEE MainJEE Main 2023Relations and Functions - Part 2
Solution:
$ 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) $
$\Rightarrow \frac{ f ( x +1)}{ f ( x )}=\frac{ f (1)}{5} $
$\Rightarrow \frac{ f (1)}{ f (0)} \cdot \frac{ f (2)}{ f (1)} \cdot \frac{ f (3)}{ f (2)}=\left(\frac{ f (1)}{5}\right)^3$
$\Rightarrow \frac{320}{5}=\frac{( f (1))^3}{5^3} \Rightarrow f (1)=20$
$\therefore 5 f ( x +1)=20 \cdot f ( x ) \Rightarrow f ( x +1)=4 f ( x )$
$\displaystyle\sum_{ n =0}^5 f ( n )=5+5.4+5.4^2+5.4^3+5.4^4+5.4^5$
$=\frac{5\left[4^6-1\right]}{3}=6825$