Question

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

• A. $\frac{7n}{2}$
• B. $\frac{7\left(n+1\right)}{2}$
• C. $7n(n+1)$
• D. $\frac{7n(n+1)}{2}$

Answer 1

The correct option is D $\frac{7n(n+1)}{2}$

$\mathrm{Given}\mathrm{that},f\left(x+y\right)=f\left(x\right)+f\left(y\right)\Rightarrow f\left(1+1\right)=f\left(1\right)+f\left(1\right)\Rightarrow f\left(2\right)=2f\left(1\right)=2\times 7f\left(2+1\right)=f\left(2\right)+f\left(1\right)\Rightarrow f\left(3\right)=2f\left(1\right)+f\left(1\right)=3f\left(1\right)=3\times 7\therefore f\left(n\right)=7n$
${\sum }_{r=1}^{n}f\left(r\right)=f\left(1\right)+f\left(2\right)+...+f\left(n\right)=7+14+21+....+7n=7\left(1+2+3+....+n\right)=\frac{7n\left(n+1\right)}{2}$