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(A) Given the functional equation $f(x+y)=f(x)+f(y)$,this is Cauchy's functional equation,which implies $f(x)=cx$ for some constant $c$.Since $f(1)=2$

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$5$

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(A) Given the functional equation $f(x+y)=f(x)+f(y)$,this is Cauchy's functional equation,which implies $f(x)=cx$ for some constant $c$.Since $f(1)=2$,we have $c(1)=2$,so $c=2$. Thus,$f(x)=2x$.Now,we calculate $g(n) = \sum_{k=1}^{n-1} f(k) = \sum_{k=1}^{n-1} 2k$.Using the sum formula $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$,we get $g(n) = 2 	imes \frac{(n-1)n}{2} = n(n-1)$.We are given $g(n)=20$,so $n(n-1)=20$.$n^2 - n - 20 = 0$.$(n-5)(n+4) = 0$.Since $n \in N$,we must have $n=5$.

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

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