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

We are given that
$f(x+y)=f\left(x\right)f\left(y\right)$ and $f\left(1\right)=2$
$\therefore f\left(2\right)=f(1+1)=f\left(1\right).f\left(1\right)=2.2=2^2$
$f\left(3\right)=f(2+1)=f\left(2\right).f\left(1\right)=2^2.2=2^3$
$f\left(4\right)=2^4$ etc.
Now $\sum _{k=1}^{n}f(a+k)=\sum _{k=1}^{n}f\left(a\right)f\left(k\right)=2^a\sum _{k=1}^{n}f\left(k\right)$
$=2^a[2+2^2+2^3+....+2^n]$
$=2^{a+1}[1+2+2^2+....+nterms]$
$=2^{a+1}\frac{1.(2^n-1)}{2-1}=2^{a+1}(2^n-1)$
$=16.(2^n-1)$ [ Given]
$\therefore 2^{a+1}=16=2^4$

$\therefore a+1=4$ or $a=3$