Mathematics - Functions Question with Solution | TestHub

From the given functional equation $f(x+y)=f(x)\cdot f(y)$, we have $f(x)=e^{kx}$. Where $k$ is a constant. Putting $x=1$, $f(1)=e^k\Rightarrow 2=e^k$. Hence from (1), $f(x)=2^x$. This can also be obtained by putting $y=1$ so that $f(x+1)=f(x)\cdot f(1)=2f(x)\Rightarrow f(x)=2f(x-1)$. Putting successively $x-1,x-2,x-3,\dots 2$ for $x$ in the above and multiplying them, we get $f(x)=2^x$.

Now,

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

$$=2^{c+1}+2^{c+2}+\dots +2^{c+n}$$

$$=2^c\cdot 2+2^c\cdot 2^2+\dots +2^c\cdot 2^n$$

$$=2^c\cdot \{2+2^2+\dots +2^n\}$$

This is a geometric progression with first term $a=2$, common ratio $r=2$, and $n$ terms.

The sum of $n$ terms is $S_n=\frac{a(r^n-1)}{r-1}$.

$$=2^c\cdot \frac{2(2^n-1)}{2-1}=2^c\cdot 2(2^n-1)$$

Given that ${\sum }_{r=1}^{n}f(c+r)=16(2^n-1)$.

So, $16(2^n-1)=2^{c+1}(2^n-1)$

$$\Rightarrow 2^{c+1}=16=2^4$$

$$\Rightarrow c+1=4$$

$$\therefore c=3$$