If f: R rightarrow R is defined as f(x+y)=f(x | vedclass

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(A) Given that $f: R \rightarrow R$ satisfies the Cauchy functional equation $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=5$. Since $f(x+y)=f(x)+

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$\frac{5 n(n+1)}{2}$

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(A) Given that $f: R ightarrow R$ satisfies the Cauchy functional equation $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=5$. Since $f(x+y)=f(x)+f(y)$,it follows that $f(n)=n \cdot f(1)$ for any integer $n$. Given $f(1)=5$,we have $f(n)=5n$. We need to find the sum $\sum_{r=1}^n f(r) = \sum_{r=1}^n 5r$. This is equal to $5 \sum_{r=1}^n r$. Using the formula for the sum of the first $n$ natural numbers,$\sum_{r=1}^n r = \frac{n(n+1)}{2}$. Therefore,$\sum_{r=1}^n f(r) = 5 	imes \frac{n(n+1)}{2} = \frac{5n(n+1)}{2}$.

If $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y)$,$\forall x, y \in R$ and $f(1)=5$,then find the value of $\sum_{r=1}^n f(r)$.

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