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

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(C) Given, $f(x+y)=f(x)+f(y)$ for all $x, y \in R$. This is Cauchy's functional equation, and for $f: R \rightarrow R$, the solution is $f(x)=cx$. Giv

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$\frac{50}{3} n(n+1)(2 n+1)$

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(C) Given, $f(x+y)=f(x)+f(y)$ for all $x, y \in R$. This is Cauchy's functional equation, and for $f: R ightarrow R$, the solution is $f(x)=cx$. Given $f(1)=10$, we have $c(1)=10$, so $c=10$. Thus, $f(x)=10x$. We need to find $\sum_{r=1}^n(f(r))^2 = \sum_{r=1}^n(10r)^2$. $= 100 \sum_{r=1}^n r^2$. Using the formula $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$, we get: $= 100 	imes \frac{n(n+1)(2n+1)}{6}$. $= \frac{50}{3} n(n+1)(2n+1)$.

If $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y), \forall x, y \in R$ and $f(1)=10$, then, $\sum_{r=1}^n(f(r))^2=$

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