If f : R → R satisfies f(x + y) = f(x) + f(y), for all x, y ∈ R and f(1) = 7, then $\sum\limits_{r=1}^{n}f(r)$ is:

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

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

$f(x+y)=f(x)+(y)⇒f(x)=ax$ $f(1)=7⇒a(1)=7$  $∴a=7$ $∴f(x)=7x⇒\sum\limits_{r=1}^{n}f(r)=7(1+2r+.....+n)=\frac{7n(n+1)}{2}$