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:
Answer & explanation
Correct answer: option 4
$f(x+y)=f(x)+(y)⇒f(x)=ax$
$f(1)=7⇒a(1)=7$ $∴a=7$
$∴f(x)=7x⇒\sum\limits_{r=1}^{n}f(r)=7(1+2r+.....+n)=\frac{7n(n+1)}{2}$