Mathematics - Functions Question with Solution | TestHub

From the given functional equation $f(x+y)=f(x).f(y)$, we have $f(x)=e^{kx}$ .......(1)

Where $k$ is a constant.

Putting $\mathrm{x}=1,\mathrm{f}(1)={\mathrm{e}}^{\mathrm{k}}\Rightarrow 2={\mathrm{e}}^{\mathrm{k}}$. Hence from (1), $\mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$.

This can also be obtained by putting $y=1$ so that

$$F(x+1)=f(x).f(1)=2f(x)\Rightarrow f(x)=2f(x-1).$$

Putting successively $\mathrm{x}-1,\mathrm{x}-2,\mathrm{x}-3,\dots 2$ for x in the above and multiplying them, we get $\mathrm{f}(\mathrm{x})=2^{\mathrm{x}}$.

Now, ${\sum }_{\mathrm{r}=1}^{\mathrm{n}}\mathrm{f}(\mathrm{c}+\mathrm{r})=\mathrm{f}(\mathrm{c}+1)+\mathrm{f}(\mathrm{c}+2)+\dots +\mathrm{f}(\mathrm{c}+\mathrm{n})$

$$\begin{array}{rlrlr}& =2^{c+1}+2^{c+2}+\dots +2^{c+n}=2^c\cdot 2+2^c\cdot 2^2+\dots +2^c\cdot 2^n& & & \end{array}$$

$$=2^c\cdot \left\{1+2+2^2+\dots tonterms\right\}$$

$$=2^c\cdot 2\cdot \frac{1\left(2^n-1\right)}{2-1}=2^c\cdot 2\left(2^n-1\right)\Rightarrow 16\left(2^n-1\right)=2^{c+1}\left(2^n-1\right)\Rightarrow 2^{c+1}=16=2^4\Rightarrow c+1=4\therefore c=3$$