If  $f:R\to R$  satisfies $f\left(x+y\right)=f\left(x\right)+f\left(y\right)$ , for all $x,y\in R$  and $f\left(1\right)=7$ , then ${\displaystyle \sum _{r=1}^{n}f\left(x\right)=}$

Put $x=1,y=0\Rightarrow f\left(0+1\right)=f\left(1\right)+f\left(0\right)\Rightarrow f\left(1\right)=f\left(1\right)+f\left(0\right)$

$\Rightarrow 7+f\left(0\right)=7\Rightarrow f\left(0\right)=0$

Put $x=1,y=1\Rightarrow f\left(1+1\right)=f\left(1\right)+f\left(1\right)\Rightarrow f\left(2\right)=7+7=14$

Put $x=2,y=1\Rightarrow f\left(2+1\right)=f\left(2\right)+f\left(1\right)\Rightarrow f\left(3\right)=14+7=21$

Put $x=3,y=1\Rightarrow f\left(3+1\right)=f\left(3\right)+f\left(1\right)\Rightarrow f\left(4\right)=21+7=28$

Put $x=n-1,y=1\Rightarrow f\left(n-1+1\right)=f\left(n-1\right)+f\left(1\right)=f\left(x\right)=72$

$\therefore {\displaystyle \sum _{r=1}^{n}f\left(r\right)}f\left(1\right)+f\left(2\right)+f\left(3\right)+\cdots +f\left(n\right)$

$=7+14+21+\cdots 7n$

$=7\left(1+2+3+\cdots +n\right)$

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