Key Concepts and Formulas

• Functional Equation Property: The functional equation $f(x+y) = f(x)f(y)$ implies that $f(x)$ is an exponential function of the form $f(x) = c^x$. Given $f(1)=2$, we deduce $f(x)=2^x$.
• Geometric Progression (GP): The sum of the first $n$ terms of a GP is given by $S_n = \frac{A(R^n - 1)}{R - 1}$, where $A$ is the first term and $R$ is the common ratio.

Step-by-Step Solution

Step 1: Determine the form of the function $f(x)$ We are given the functional equation $f(x+y) = f(x)f(y)$ for all natural numbers $x, y$, and $f(1) = 2$. Let $x=1, y=1$. Then $f(1+1) = f(1)f(1)$, which means $f(2) = (f(1))^2 = 2^2 = 4$. Let $x=2, y=1$. Then $f(2+1) = f(2)f(1)$, which means $f(3) = 4 \times 2 = 2^3 = 8$. Continuing this pattern, we can infer that for any natural number $n$, $f(n) = 2^n$. We can prove this by induction. Base case: $f(1) = 2^1 = 2$, which is given. Inductive step: Assume $f(m) = 2^m$ for some natural number $m$. Then $f(m+1) = f(m)f(1)$ (from the functional equation) $= 2^m \times 2 = 2^{m+1}$. Thus, by induction, $f(x) = 2^x$ for all natural numbers $x$.

Step 2: Analyze the given summation We are given the equation $\sum\limits_{k = 1}^{10} {f(a + k) = 16\left( {{2^{10}} - 1} \right)}$. Substitute $f(x) = 2^x$ into the summation: $\sum\limits_{k = 1}^{10} {2^{a + k}} = 16\left( {{2^{10}} - 1} \right)$

Step 3: Expand and simplify the summation The summation can be written as: $2^{a+1} + 2^{a+2} + 2^{a+3} + \dots + 2^{a+10}$ We can factor out $2^a$ from each term: $2^a (2^1 + 2^2 + 2^3 + \dots + 2^{10})$ The expression in the parenthesis is a geometric progression with the first term $A = 2^1 = 2$, the common ratio $R = 2$, and the number of terms $n = 10$. Using the formula for the sum of a GP, $S_n = \frac{A(R^n - 1)}{R - 1}$: $2^1 + 2^2 + \dots + 2^{10} = \frac{2(2^{10} - 1)}{2 - 1} = 2(2^{10} - 1)$ So, the summation becomes: $2^a \left[ 2(2^{10} - 1) \right] = 2^{a+1}(2^{10} - 1)$

Step 4: Equate the simplified summation with the given right-hand side We have $2^{a+1}(2^{10} - 1) = 16(2^{10} - 1)$. Since $2^{10} - 1 \neq 0$, we can divide both sides by $(2^{10} - 1)$: $2^{a+1} = 16$ We know that $16 = 2^4$. Therefore, $2^{a+1} = 2^4$.

Step 5: Solve for 'a' Equating the exponents: $a+1 = 4$ $a = 4 - 1$ $a = 3$

Common Mistakes & Tips

• Incorrectly identifying $f(x)$: Ensure that the functional equation and initial condition are correctly used to derive the explicit form of $f(x)$.
• Error in GP sum formula: Double-check the formula for the sum of a geometric progression, especially the first term, common ratio, and number of terms.
• Algebraic mistakes: Be careful with exponent rules and algebraic manipulations when simplifying the summation and solving for 'a'.

Summary

The problem requires understanding the properties of functional equations and geometric progressions. First, we deduced that $f(x) = 2^x$ from the given functional equation and initial condition. Then, we expressed the summation $\sum\limits_{k = 1}^{10} {f(a + k)}$ as a geometric series. By simplifying this series and equating it to the given expression $16(2^{10} - 1)$, we were able to solve for the unknown natural number 'a'. The derived value of 'a' is 3.

The final answer is \boxed{3}.