A Function All The Way

We know $f(x+y)=f(x)\times f(y)$ and $f(1)=2$

$\Rightarrow f(x)=f(1+1+1...$ $x$ $times)$

$\Rightarrow f(x)=[f(1)]^x$

$\Rightarrow f(x)=2^x$ , since $f(1)=2$ .

Therefore $f(a+k)=2^a\times2^k$

$\large\Rightarrow \sum_{k=1}^{n} f(a+k)=2^a\times2^1+2^a\times2^2+2^a\times2^3 .......2^a\times2^n$

$\large\Rightarrow \sum_{k=1}^{n} f(a+k)=\large\frac{2^a[2(1-2^n)]}{1-2}$ using the formula of G.P.

$\large\Rightarrow \sum_{k=1}^{n} f(a+k)=\large\frac{[2^{a+1}-2^{n+a+1})]}{-1}$ , which is equal to $16(2^n-1)$ , according to the question.

Therefore $\large\frac{[2^{a+1}-2^{n+a+1})]}{-1}=16(2^n-1)$

Simplyfying we get $2^{a+1}=16$

which gives $a=3$ .

$\large\frac{a^2}{f(a)}=\frac{3^2}{2^3}=\frac{9}{8}$

So, the answer is $9+8=\boxed{17}$

It appears that $f(n)=2^n$ for the first few $n$ . Let us prove that the claim is true for all $n \ge 1$ .

Proof: For $n=1$ , $f(1) = 2^1 = 2$ as given, therefore, the claim is true for $n=1$ . Now, assuming the claim is true for $n$ , then $f({\color{#D61F06}n+1}) = f(1)f(n) = 2\times 2^n = 2^{\color{#D61F06}n+1}$ , therefore, the claim is also true for $n+1$ hence it is true for all $n \ge 1$ .

Now, we have $\displaystyle\sum_{k=1}^n f(a+k) = \sum_{k=1}^n f(a)f(k) = f(a) \sum_{k=1}^n f(k) = 2^a \sum_{k=1}^n 2^k = 2^{a+1} \left(\frac {2^n-1}{2-1}\right) = 2^{a+1} \left(2^n-1 \right)$ .

Therefore, $2^{a+1} = 16 = 2^4 \implies a+1=4 \implies a = \boxed{3}$ .