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Step-by-Step Solution
Step 1: Understand the given functional equation and initial condition
We are told that the function $f$ satisfies:
1) $f(x + y) = f(x)\,f(y)$ for all natural numbers $x$ and $y$.
2) $f(1) = 2$.
From this, we can identify that $f(x)$ has the form of an exponential function.
Step 2: Determine the explicit form of $f(x)$
Given $f(1) = 2$, we check:
• $f(2) = f(1 + 1) = f(1)\,f(1) = 2 \times 2 = 2^2$.
• $f(3) = f(2 + 1) = f(2)\,f(1) = 2^2 \times 2 = 2^3$.
By induction, we see that for any natural number $n$, $f(n) = 2^n$.
Step 3: Write down the summation in terms of powers of 2
We want to evaluate
$\displaystyle \sum_{k = 1}^{10} f(a + k) = 16 \bigl(2^{10} - 1 \bigr).$
Since $f(a + k) = 2^{a + k}$, the sum becomes
$\displaystyle \sum_{k = 1}^{10} 2^{a + k}.$
Explicitly, this is
$\displaystyle 2^{a+1} + 2^{a+2} + \dots + 2^{a+10}.$
Step 4: Recognize the geometric series and sum it
The above series is a geometric series with the first term $2^{a+1}$ and common ratio $2$. Its 10 terms add up to:
$\displaystyle 2^{a+1} + 2^{a+2} + \dots + 2^{a+10}
= 2^a \left(2^1 + 2^2 + \dots + 2^{10}\right).$
Now, the sum of powers of 2 from $2^1$ to $2^{10}$ is:
$\displaystyle 2^1 + 2^2 + \dots + 2^{10}
= 2(1 + 2 + 2^2 + \dots + 2^9).$
Alternatively, use the standard geometric series formula:
$\displaystyle 2 + 2^2 + \dots + 2^{10}
= 2\,\frac{2^{10} - 1}{2 - 1}
= 2(2^{10} - 1).$
Step 5: Set up the equation for the given sum
Hence,
$\displaystyle \sum_{k = 1}^{10} 2^{a + k}
= 2^a \times \left[2(2^{10} - 1)\right]
= 16 (2^{10} - 1).$
So we need
$\displaystyle 2^a \times 2 \bigl(2^{10} - 1 \bigr)
= 16 (2^{10} - 1).$
Step 6: Solve for $a$
Divide both sides by $(2^{10} - 1)$ (which is nonzero):
$\displaystyle 2^a \times 2 = 16.$
Thus,
$\displaystyle 2^{a + 1} = 2^4 \quad \Rightarrow \quad a + 1 = 4 \quad \Rightarrow \quad a = 3.$
Step 7: State the final answer
Therefore, the natural number $a$ that satisfies the given conditions is
$\boxed{3}.$
