Mathematics - Functions Question with Solution | TestHub

Given, equation is:

$5f\left(x+y\right)=f\left(x\right)·f\left(y\right) ...\left(1\right)$

At $x=0, y=0$, we get

$5f\left(0\right)=f{\left(0\right)}^2\Rightarrow f\left(0\right)=5$

Put $y=1$ in $\left(1\right)$, then we get

$5f\left(x+1\right)=f\left(x\right)·f\left(1\right)$

$\Rightarrow \frac{f\left(x+1\right)}{f\left(x\right)}=\frac{f\left(1\right)}{5}$

$\Rightarrow \frac{f\left(1\right)}{f\left(0\right)}·\frac{f\left(2\right)}{f\left(1\right)}·\frac{f\left(3\right)}{f\left(2\right)}={\left(\frac{f\left(1\right)}{5}\right)}^3$

$\Rightarrow \frac{320}{5}=\frac{{\left(f\left(1\right)\right)}^3}{5^3}\Rightarrow f\left(1\right)=20$

$\therefore 5f\left(x+1\right)=20·f\left(x\right)$

$\Rightarrow f\left(x+1\right)=4f\left(x\right)$

So,

$$\begin{gathered} f\left(1\right)=4f\left(0\right)=4·5 \\ f\left(2\right)=4f\left(1\right)=4^2·5 \\ f\left(3\right)=4f\left(2\right)=4^3·5 \\ f\left(4\right)=4f\left(3\right)=4^4·5 \\ f\left(5\right)=4f\left(4\right)=4^5·5 \end{gathered}$$

$\sum _{n=0}^{5}f\left(n\right)=5+5·4+5·4^2+5·4^3+5·4^4+5·4^5$

$\Rightarrow \sum _{n=0}^{5}f\left(n\right)=\frac{5\left(4^6-1\right)}{3}=6825$