Key Concepts and Formulas

• Cauchy's Functional Equation: For a function $f: \mathbb{R} \to \mathbb{R}$, if $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, then for any integer $n$, $f(nx) = nf(x)$. For $x=1$, this implies $f(n) = nf(1)$.
• Sum of the first $m$ natural numbers: The sum of the first $m$ positive integers is given by the formula $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$.
• Quadratic Equations: A quadratic equation of the form $ax^2 + bx + c = 0$ can be solved by factoring, completing the square, or using the quadratic formula.

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Step-by-Step Solution

Step 1: Determine the explicit form of $f(k)$ for positive integers $k$. We are given the functional equation $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$, and $f(1) = 2$. Using the property of Cauchy's functional equation for integers, we can establish a pattern for $f(k)$. For $k=1$, $f(1) = 1 \cdot f(1) = 1 \cdot 2 = 2$. For $k=2$, $f(2) = f(1+1) = f(1) + f(1) = 2 + 2 = 4$. This can also be written as $f(2) = 2 \cdot f(1) = 2 \cdot 2 = 4$. For $k=3$, $f(3) = f(2+1) = f(2) + f(1) = 4 + 2 = 6$. This can also be written as $f(3) = 3 \cdot f(1) = 3 \cdot 2 = 6$. By induction or by direct application of $f(nx) = nf(x)$ with $x=1$, we can conclude that for any positive integer $k$, $f(k) = k \cdot f(1)$. Given $f(1) = 2$, we have $f(k) = k \cdot 2 = 2k$.

Step 2: Express $g(n)$ as a sum and simplify it. The function $g(n)$ is defined as $g(n) = \sum_{k=1}^{(n-1)} f(k)$ for $n \in \mathbb{N}$. Substituting the expression for $f(k)$ found in Step 1, we get: $g(n) = \sum_{k=1}^{(n-1)} 2k$ We can factor out the constant 2 from the summation: $g(n) = 2 \sum_{k=1}^{(n-1)} k$ The sum $\sum_{k=1}^{(n-1)} k$ is the sum of the first $(n-1)$ natural numbers. Using the formula for the sum of the first $m$ natural numbers, where $m = n-1$: $\sum_{k=1}^{(n-1)} k = \frac{(n-1)((n-1)+1)}{2} = \frac{(n-1)n}{2}$ Substituting this back into the expression for $g(n)$: $g(n) = 2 \left( \frac{(n-1)n}{2} \right)$ $g(n) = n(n-1)$ $g(n) = n^2 - n$

Step 3: Solve the equation for $n$ given $g(n) = 20$. We are given that $g(n) = 20$. We set our simplified expression for $g(n)$ equal to 20: $n^2 - n = 20$ To solve this equation, we rearrange it into a standard quadratic form by moving all terms to one side: $n^2 - n - 20 = 0$ We can solve this quadratic equation by factoring. We look for two numbers that multiply to -20 and add up to -1. These numbers are -5 and 4. Therefore, we can factor the equation as: $(n - 5)(n + 4) = 0$ This equation yields two possible solutions for $n$:

1. $n - 5 = 0 \implies n = 5$
2. $n + 4 = 0 \implies n = -4$

Step 4: Validate the solution based on the domain of $n$. The problem states that $n \in \mathbb{N}$, which means $n$ must be a natural number (a positive integer: 1, 2, 3, ...). Let's check our two solutions:

• $n = 5$: This is a positive integer and thus a natural number.
• $n = -4$: This is a negative integer and therefore not a natural number.

Since $n$ must be a natural number, we discard the solution $n = -4$. The only valid solution is $n = 5$.

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Common Mistakes & Tips

• Domain of $n$: Always pay close attention to the domain specified for variables like $n$. In this case, $n \in \mathbb{N}$ was crucial for selecting the correct solution from the quadratic equation.
• Cauchy's Equation for Integers: Recognize that $f(x+y)=f(x)+f(y)$ implies $f(n)=nf(1)$ for integer $n$. This is a direct and powerful consequence.
• Summation Formula: Ensure you correctly identify the upper limit of the summation when applying the sum of natural numbers formula. Here, it was $(n-1)$, not $n$.

Summary The problem required us to first determine the explicit form of the function $f(k)$ for positive integers $k$, which we found to be $f(k) = 2k$ by utilizing the given Cauchy's functional equation and the value of $f(1)$. We then substituted this into the definition of $g(n)$ to obtain a sum of an arithmetic progression, which simplified to $g(n) = n(n-1)$. Finally, we solved the equation $n(n-1) = 20$ for $n$, and after considering the constraint that $n$ must be a natural number, we found the unique solution $n=5$.

The final answer is $\boxed{5}$ which corresponds to option (C).