Question

Find the natural number $a$ for which ${\sum }_{k=1}^{n}f(a+k)=16(2^n-1)$ where the function $f$ satisfies the relation $f(x+y)=f\left(x\right).f\left(y\right)$ for all natural numbers $x,y$ and further $f\left(1\right)=2$.

Answer 1

$f(x+y)=f\left(x\right).f\left(y\right),f\left(1\right)=2$
${\sum }_{k=1}^{n}f(a+k)=16(2^n-1)$
$\Rightarrow f\left(a\right).f\left(1\right)+f\left(a\right).f\left(2\right)+...+f\left(a\right).f\left(n\right)=16(2^n-1)$
$\Rightarrow f\left(a\right).\left[f\right(1)+f(2)+...+f(n\left)\right]=16(2^n-1)$
Also, $f\left(2\right)=f(1+1)=f\left(1\right).f\left(1\right)=\left[f\right(1){]}^2=4$
Similarly, $f\left(3\right)=f(2+1)=f\left(2\right).f\left(1\right)=4\times 2=8$
$\therefore f\left(a\right).\left[f\right(1)+f(2)+...+f(n\left)\right]=16(2^n-1)$ becomes $f\left(a\right).[2+4+8+...+2^n]=16(2^n-1)$
$\Rightarrow f\left(a\right).\left[2\right(2^n-1\left)\right]=16(2^n-1)$
Thus, $f\left(a\right)=8$ which implies $a=3$