Question

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

Answer 1

It is given that,
$f(x+y)=f\left(x\right)f\left(y\right)$ and $f\left(1\right)=2$
$\Rightarrow f(1+1)=f\left(1\right)f\left(1\right)$
$\therefore f\left(2\right)=2.2=2^2$
$\Rightarrow f(1+2)=f\left(1\right)f\left(2\right)$
$\therefore f\left(3\right)={2.2}^2=2^3$
$\Rightarrow f(2+2)=f\left(2\right)f\left(2\right)$
$\therefore f\left(4\right)=2^2{.2}^2=2^4$
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Similarly, $f\left(k\right)=2^k$ and $f\left(a\right)=2^a$
Hence, ${\sum }_{k=1}^{n}f(a+k)={\sum }_{k=1}^{n}f\left(a\right)f\left(k\right)$
$=f\left(a\right){\sum }_{k=1}^{k}f\left(k\right)=2^a{\sum }_{k=1}^k{2}^k$
$=2^a\{2^1+2^2+2^3+.....+2^n\}$
$=2^a\left\{\frac{2.(2^n-1)}{2-1}\right\}$
But, ${\sum }_{k=1}^{n}f(a+k)=16(2^n-1)$
$\therefore 2^{a+1}(2^n-1)=16(2^n-1)$
$\Rightarrow 2^{a+1}=2^4$
$\Rightarrow a+1=4$
$\therefore a=3$