1. We are given $f(x+y)=f(x)f(y) \quad \text{for all natural numbers } x,y,$ and $f(1)=2.$

We need to find the natural number $a$ such that $\sum_{k=1}^{10} f(a+k)=16(2^{10}-1).$

2. First, determine the form of $f(n)$ for natural numbers $n$.

Since $f(x+1)=f(x)f(1)=2f(x),$ starting from $f(1)=2$, we get: $f(2)=f(1+1)=f(1)f(1)=2\cdot 2=2^2,$ $f(3)=f(2+1)=f(2)f(1)=2^2\cdot 2=2^3,$ and in general, $f(n)=2^n \quad \text{for all natural } n.$

3. Substitute this into the given sum: $\sum_{k=1}^{10} f(a+k)=\sum_{k=1}^{10} 2^{a+k}.$

Factor out $2^a$: $\sum_{k=1}^{10} 2^{a+k}=2^a\sum_{k=1}^{10}2^k.$

Now, $\sum_{k=1}^{10}2^k=2+2^2+\cdots+2^{10}=2(2^{10}-1).$

So the left-hand side becomes $2^a \cdot 2(2^{10}-1)=2^{a+1}(2^{10}-1).$

4. Given that this equals $16(2^{10}-1),$ we compare: $2^{a+1}(2^{10}-1)=16(2^{10}-1).$

Since $(2^{10}-1)\neq 0$, cancel it: $2^{a+1}=16=2^4.$

Therefore, $a+1=4 \implies a=3.$

5. Check options:

• A: $2$ ❌
• B: $16$ ❌
• C: $4$ ❌
• D: $3$ ✅

Hence the correct answer is D.