Step-by-Step Solution

Given: The function f satisfies f(x + y) = f(x) f(y) for all natural numbers x, y, and f(1) = 2. Also, the sum is given as:

$\sum _{k=1}^{10}f(a+k)=16(2^{10}-1)$

We need to find the natural number 'a'.

Step 1: Understand the Functional Equation

The equation f(x + y) = f(x) f(y) is a standard form of an exponential function. Given f(1) = 2, we can determine the general form of f(x).

Let x and y be natural numbers. Since f(1) = 2, we can find f(2):

$f\left(2\right)=f(1+1)=f\left(1\right)f\left(1\right)=2\times 2=2^2$

Similarly, f(3) = f(2+1) = f(2) f(1) = 2² × 2 = 2³, and so on. In general, for any natural number n:

$f\left(n\right)=2^n$

Step 2: Express the Given Sum in Terms of f

The sum is: $\sum _{k=1}^{10}f(a+k)$

Using the general form f(n) = 2ⁿ, we can write:

$f(a+k)=2^{a+k}=2^a·2^k$

So the sum becomes:

$\sum _{k=1}^{10}{2}^a·2^k=2^a\sum _{k=1}^{10}{2}^k$

Step 3: Evaluate the Summation

The sum $\sum _{k=1}^{10}{2}^k$ is a geometric series with first term 2 and common ratio 2.

Sum of geometric series: $S=a\frac{r^n-1}{r-1}$, where a is first term, r is common ratio, n is number of terms.

Here, a = 2, r = 2, n = 10:

$S=2·\frac{2^{10}-1}{2-1}=2·(2^{10}-1)$

So the total sum is:

$2^a·2(2^{10}-1)=2^{a+1}(2^{10}-1)$

Step 4: Equate to the Given Value and Solve for 'a'

This sum is given to be equal to 16(2¹⁰ - 1):

$2^{a+1}(2^{10}-1)=16(2^{10}-1)$

Since 2¹⁰ - 1 ≠ 0, we can cancel it from both sides:

$2^{a+1}=16$

Now, 16 = 2⁴, so:

$2^{a+1}=2^4$

Therefore, the exponents must be equal:

$a+1=4$

$a=3$

Final Answer

The natural number 'a' is 3.

Related Topics

• Functional Equations
• Exponential Functions
• Geometric Series
• Summation Notation

Key Formulae and Theory

Exponential Functional Equation: If f(x+y) = f(x)f(y) for all x, y, and f(1) = c, then f(n) = cⁿ for natural numbers n.

Geometric Series Sum: The sum of the first n terms of a geometric series with first term a and common ratio r is:

$S=a\frac{r^n-1}{r-1}$ for r ≠ 1.

Exponent Rules: aᵐ ⋅ aⁿ = aᵐ⁺ⁿ and (aᵐ)ⁿ = aᵐⁿ.