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If $\mathrm{f}: \mathbb{R} \rightarrow \mathbb{R}$ is defined by $\mathrm{f}(\mathrm{x}+\mathrm{y})=\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \forall \mathrm{x}, \mathrm{y} \in$ $\mathbb{R}$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$
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The correct answer is:
$\frac{7 n(n+1)}{2}$
Since $f(x+y)=f(x)+f(y), \forall x, y \in R$
$\Rightarrow f(x)=a x$ where $a \in R$
Since given, $\mathrm{f}(1)=7 \Rightarrow 7=\mathrm{a} \times 1 \Rightarrow \mathrm{a}=7$
$\Rightarrow \mathrm{f}(\mathrm{x})=7 \mathrm{x}$
Now, $\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{r})=\sum_{\mathrm{r}=1}^{\mathrm{n}} 7 \mathrm{r}=\frac{7 \mathrm{n}(\mathrm{n}+1)}{2}$
$\Rightarrow f(x)=a x$ where $a \in R$
Since given, $\mathrm{f}(1)=7 \Rightarrow 7=\mathrm{a} \times 1 \Rightarrow \mathrm{a}=7$
$\Rightarrow \mathrm{f}(\mathrm{x})=7 \mathrm{x}$
Now, $\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{f}(\mathrm{r})=\sum_{\mathrm{r}=1}^{\mathrm{n}} 7 \mathrm{r}=\frac{7 \mathrm{n}(\mathrm{n}+1)}{2}$
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