<s>[INST] Find the natural number $a$ for which $\sum^n_{k=1}f(a+k)=16(2^n-1)$, given that $f(x+y)=f(x)\,f(y)$. 
Find the natural number $a$ for which
  $$\sum^n_{k=1}f(a+k)=16(2^n-1)\,,$$
  where the function $f:\mathbb{N}\to\mathbb{N}$ satisfies the relation 
  $$f(x+y)=f(x)f(y)\text{ for all }x,y\in \mathbb{N}$$
  and further $f(1) = 2$.

Please solve this.
 [/INST] The first thing to find is what $f(x)$ is
$$f(1) = 2 = f(1+0) = f(1)*f(0)$$
Therefore, $f(0) = 1$
$$f(2) = f(1)f(1) = 2^2$$
$$f(3) = f(2)f(1) = f(1)f(1)f(1) = 2^3$$
Clearly, $f(x) = 2^x, x \in \mathbb{N}$
Now 
$$\sum_{k=1}^n 2^{a+k} = 16(2^n-1)$$
$$2^a\sum_{k=1}^n 2^k = 16(2^n-1)$$
$$2^a \cdot \frac{2^{n+1} - 2}{2-1} = 16(2^n-1)$$
$$2^a(2 \cdot(2^n-1)) = 16(2^n-1)$$
Clearly $a = 3$
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