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(C) Given $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation on $\mathbb{N}$,which implies $f(n) = cn$ for some constant $c$.Since $f(1) =

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$10$

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(C) Given $f(x + y) = f(x) + f(y)$,this is Cauchy's functional equation on $\mathbb{N}$,which implies $f(n) = cn$ for some constant $c$.Since $f(1) = \frac{1}{5}$,we have $c(1) = \frac{1}{5}$,so $f(n) = \frac{n}{5}$.Now,substitute $f(n) = \frac{n}{5}$ into the summation:$\sum_{n=1}^m \frac{n/5}{n(n+1)(n+2)} = \frac{1}{5} \sum_{n=1}^m \frac{1}{(n+1)(n+2)}$.Using partial fractions,$\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}$.So,the sum becomes $\frac{1}{5} \sum_{n=1}^m \left( \frac{1}{n+1} - \frac{1}{n+2} \right)$.This is a telescoping sum:$\frac{1}{5} \left( (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{m+1} - \frac{1}{m+2}) \right) = \frac{1}{5} \left( \frac{1}{2} - \frac{1}{m+2} \right)$.Given the sum is $\frac{1}{12}$,we have $\frac{1}{5} \left( \frac{m+2-2}{2(m+2)} \right) = \frac{1}{12} \implies \frac{m}{10(m+2)} = \frac{1}{12}$.$12m = 10m + 20 \implies 2m = 20 \implies m = 10$.

Suppose $f$ is a function satisfying $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$ and $f(1) = \frac{1}{5}$. If $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}$,then $m$ is equal to $...............$.

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