If f:R to R satisfies f(x + y) = f(x) + f(y) | vedclass

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(D) 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

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$\frac{7n(n + 1)}{2}$

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(D) 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 $f(1) = 7$,we have $c(1) = 7$,so $c = 7$. Thus,$f(x) = 7x$. Now,we need to calculate the sum $\sum_{r = 1}^n f(r) = \sum_{r = 1}^n 7r$. This simplifies to $7 \sum_{r = 1}^n r$. Using the formula for the sum of the first $n$ natural numbers,$\sum_{r = 1}^n r = \frac{n(n + 1)}{2}$. Therefore,$\sum_{r = 1}^n f(r) = 7 	imes \frac{n(n + 1)}{2} = \frac{7n(n + 1)}{2}$.

If $f:R \to R$ satisfies $f(x + y) = f(x) + f(y)$ for all $x, y \in R$ and $f(1) = 7$,then $\sum_{r = 1}^n f(r)$ is

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