Q. If $f:R \rightarrow R$ satisfies $f(x + y) = f(x) + f(y)$, for all $x,\ y \in R$ and $f(1) = 7$, then $\sum_{r = 1}^{n}{f(r)}$ is
A) $\frac{7n}{2}$
B) $\frac{7(n + 1)}{2}$
C) $7n(n + 1)$
D) $\frac{7n(n + 1)}{2}$
Explanation
(d)
$\ \sum_{r = 1}^{n}{f(r) = f(1) + f(2) + f(3) + \ldots + f(n)}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = f(1) + 2f(1) + 3f(1) + ...nf(n)$
[since, $f(x + y) = f(x) + f(y)$]
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = (1 + 2 + 3 + ... + n)f(1) = f(1)\sum_{}^{}n$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{7n(n + 1)}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because f(1) = 7\ (given)\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.