Question:

Step 1: Determine the functional form of $f(k)$ for natural numbers. The function $f: R \to R$ satisfies Cauchy's functional equation, $f(x+y) = f(x) + f(y)$. We are given $f(1)=2$. For any natural number $k$, we can show by induction that $f(k) = k \cdot f(1)$. Using the given value $f(1)=2$: $$f(k) = 2k$$ Step 2: Express $g(n)$ using the determined form of $f(k)$. The function $g(n)$ is defined as a sum: $$g(n) = \sum_{k=1}^{n-1} f(k)$$ Substitute the expression $f(k) = 2k$ into the sum: $$g(n) = \sum_{k=1}^{n-1} 2k$$ Step 3: Evaluate the summation for $g(n)$. We can factor out the constant 2 from the summation and use the formula for the sum of the first $m$ natural numbers, $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$. In this case, $m = n-1$. $$g(n) = 2 \sum_{k=1}^{n-1} k$$ $$g(n) = 2 \cdot \frac{(n-1)((n-1)+1)}{2}$$ $$g(n) = 2 \cdot \frac{(n-1)n}{2}$$ $$g(n) = n(n-1)$$ Step 4: Solve for $n$ using the given condition. We are given that $g(n) = 20$. Equate the derived expression for $g(n)$ to 20: $$n(n-1) = 20$$ Expand and rearrange the equation to form a quadratic equation: $$n^2 - n - 20 = 0$$ Factor the quadratic equation: $$(n-5)(n+4) = 0$$ This gives two possible values for $n$: $$n=5 \quad \text{or} \quad n=-4$$ Step 5: Select the valid value for $n$. In the context of the sum $g(n) = \sum_{k=1}^{n-1} f(k)$, $n$ must be a positive integer, as $k$ ranges from $1$ to $n-1$. If $n=1$, the sum is empty, $g(1)=0$. If $n>1$, $n-1$ is a positive integer. Therefore, $n=5$ is the only valid solution. The final answer is $\boxed{5}$.