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(C) Given the Cauchy functional equation $f(x+y) = f(x) + f(y)$,the solution is of the form $f(x) = cx$ for some constant $c$. Since $f(1) = 7$,we hav

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$63$

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(C) Given the Cauchy functional equation $f(x+y) = f(x) + f(y)$,the solution is of the form $f(x) = cx$ for some constant $c$. Since $f(1) = 7$,we have $c(1) = 7$,which implies $c = 7$. Thus,$f(x) = 7x$. Now,we are given the sum $\sum_{r=1}^n f(r) = 14112$. Substituting $f(r) = 7r$,we get $\sum_{r=1}^n 7r = 14112$. Factoring out the constant $7$,we have $7 \sum_{r=1}^n r = 14112$. Using the sum formula $\sum_{r=1}^n r = \frac{n(n+1)}{2}$,we get $7 	imes \frac{n(n+1)}{2} = 14112$. Dividing both sides by $7$,we get $\frac{n(n+1)}{2} = 2016$. Multiplying by $2$,we get $n(n+1) = 4032$. Since $63 	imes 64 = 4032$,we have $n(n+1) = 63 	imes 64$. Therefore,$n = 63$.

If $f: \mathbb{R} \rightarrow \mathbb{R}$ is such that $f(x+y) = f(x) + f(y)$ for all $x, y \in \mathbb{R}$,$f(1) = 7$,and $\sum_{r=1}^n f(r) = 14112$,then $n = $

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