Question

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

• A. 6575
• B. 6825
• C. 6875
• D. 6525

Answer 1

Answer: (B)

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