Let $$f : R \to R$$ be a function which satisfies $$f(x + y) = f(x) + f(y)$$, $$\forall x, y \in R$$. If $$f(1) = 2$$ and $$g(n) = \sum_{k=1}^{(n-1)} f(k)$$, $$n \in N$$ then the value of $$n$$, for which $$g(n) = 20$$, is:

We have a real-valued function that obeys the functional equation $$f(x+y)=f(x)+f(y) \quad \forall\,x,y\in\mathbb R.$$

First we find the values of $$f$$ at positive integers. Put $$y=1$$ and $$x=k-1$$ (where $$k\in\mathbb N$$) in the given relation:

$$f(k)=f\big((k-1)+1\big)=f(k-1)+f(1).$$

Because $$f(1)=2,$$ this becomes

$$f(k)=f(k-1)+2.$$

Apply the same step repeatedly (mathematical induction):

For $$k=2$$: $$f(2)=f(1)+2=2+2=4.$$

For $$k=3$$: $$f(3)=f(2)+2=4+2=6.$$

We notice the pattern $$f(k)=2k.$$ We justify it formally by induction: assume $$f(m)=2m$$ for some integer $$m\ge 1,$$ then

$$f(m+1)=f(m)+2=2m+2=2(m+1),$$

which completes the proof. Hence for every natural number $$k$$

$$f(k)=2k.$$

Now we examine the auxiliary function $$g(n)$$ defined by the sum

$$g(n)=\sum_{k=1}^{\,n-1} f(k), \quad n\in\mathbb N.$$

Substituting $$f(k)=2k$$ gives

$$g(n)=\sum_{k=1}^{\,n-1} 2k = 2\sum_{k=1}^{\,n-1} k.$$

We state the standard formula for the sum of the first $$m$$ natural numbers: $$\sum_{k=1}^{m} k = \dfrac{m(m+1)}{2}.$$ Here $$m=n-1,$$ so

$$\sum_{k=1}^{\,n-1} k = \dfrac{(n-1)n}{2}.$$

Substituting this into the expression for $$g(n):$$

$$g(n)=2\left(\dfrac{(n-1)n}{2}\right)=n(n-1).$$

The problem states that $$g(n)=20.$$ Therefore we must solve

$$n(n-1)=20.$$

Expanding gives the quadratic equation

$$n^{2}-n-20=0.$$

Factorising, we write

$$(n-5)(n+4)=0.$$

This yields two roots: $$n=5$$ and $$n=-4.$$ Because $$n$$ is a natural number, we discard the negative value and keep

$$n=5.$$

Hence, the correct answer is Option A.