If f(x + y) = f(x)f(y) and Σ(x=1 to ∞) f(x) = 2, x, y ∈ N where N is the set of all natural numbers,

Q: Why can we assume $f(x) = a^x$?

The functional equation $f(x + y) = f(x)f(y)$ is the defining property of exponential functions. For natural numbers, this implies $f(1) = a$, $f(2) = f(1+1) = f(1)f(1) = a^2$, and by induction, $f(x) = a^x$. It is a powerful tool that turns complex functional relationships into simple algebraic ones.

Q: Why is the sum $\frac{a}{1-a} = 2$?

The series is $a + a^2 + a^3 + \dots = 2$. This is a geometric progression with first term $A=a$ and common ratio $r=a$. The sum formula $S = \frac{A}{1-r}$ is valid only when $|r| < 1$. Since $a = 2/3$, which is less than 1, the formula is perfectly applicable.

Q: Can I just solve for $f(4)$ and $f(2)$ directly?

You could, but it's inefficient. By simplifying $\frac{f(4)}{f(2)}$ to $a^2$ first, you avoid calculating large powers and then dividing. Always look for algebraic simplifications before plugging in numbers.

Question

If $f(x + y) = f(x)f(y)$ and $\sum_{x=1}^{\infty} f(x) = 2$, $x, y \in \mathbb{N}$ where $\mathbb{N}$ is the set of all natural numbers, then the value of $\frac{f(4)}{f(2)}$ is:

Options:

(A) $\frac{1}{9}$

(B) $\frac{4}{9}$

(C) $\frac{1}{3}$

(D) $\frac{2}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Functional Equation

Imagine you are standing before a locked door, and the key is a simple, elegant equation: $f(x + y) = f(x)f(y)$. This is a fundamental signature in mathematics.

Whenever you encounter a function that turns the addition of inputs into the multiplication of outputs, your mind should immediately leap to the world of exponential functions. This is because exponents possess the property:
$$a^{x+y} = a^x \cdot a^y$$

The domain is restricted to natural numbers, $x, y \in \mathbb{N}$, which means we are dealing with discrete steps. Let us define $f(1) = a$.

Then, $f(2) = f(1+1) = f(1)f(1) = a \cdot a = a^2$. Following this logic, $f(3) = f(2+1) = f(2)f(1) = a^2 \cdot a = a^3$.

By induction, we can confidently state that for any natural number $x$:
$$f(x) = a^x$$

### The Infinite Series

Now, let us turn our attention to the second piece of the puzzle: $\sum_{x=1}^{\infty} f(x) = 2$. Substituting our generalized function $f(x) = a^x$, the series becomes:
$$a^1 + a^2 + a^3 + \dots = 2$$

This is a classic infinite geometric progression (G.P.). For an infinite G.P. with first term $A$ and common ratio $r$, the sum is $S_{\infty} = \frac{A}{1-r}$, provided $|r| < 1$.

In our specific series, the first term $A = a$ and the common ratio $r = a$. Therefore, the sum is:
$$\frac{a}{1-a} = 2$$

Let us solve for $a$. Cross-multiplying gives $a = 2(1 - a)$, which simplifies to $a = 2 - 2a$. Adding $2a$ to both sides, we get $3a = 2$, or:
$$a = \frac{2}{3}$$

### The Final Calculation

We are almost there. The question asks for the value of $\frac{f(4)}{f(2)}$. Instead of calculating $f(4)$ and $f(2)$ separately, let us use the power of algebra.

Since $f(x) = a^x$, the ratio becomes $\frac{a^4}{a^2}$. Using the laws of exponents, this simplifies beautifully to:
$$a^{4-2} = a^2$$

Now, we simply substitute our value $a = \frac{2}{3}$ into this expression:
$$a^2 = \left( \frac{2}{3} \right)^2 = \frac{4}{9}$$

The elegance of this cancellation is the reward for our patience. We have navigated the functional equation, tamed the infinite series, and arrived at the final result of **$\frac{4}{9}$** with precision and clarity.

Animated Solution for Mathematics - Sequence and Series: If $f (x + y) = f (x) f (y)$ and $\sum_{x = 1}^{\infty} f (x) = 2$ , $x, y \in N$ where $N$ is the set of all natural numbers, then the value of $\frac{f ( 4 )}{f ( 2 )}$ is:

Select Answer:

Visualized Solution

Analyze the Functional Equation

Generalize the Function $f (x)$

Set up the Infinite Series

Apply the Infinite G.P. Formula

Solve for the Constant $a$

Express the Required Ratio

Final Calculation

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Functional Equation

The Infinite Series

The Final Calculation

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