Question

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 $\sum _{r=1}^{n}f\left(r\right)$

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

Answer 1

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

$\because f\left(1\right)=7$

$\therefore f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)=7+7=14$

Similarly $f\left(3\right)=f(1+2)=f\left(1\right)+f\left(2\right)=7+14=21$

$f\left(4\right)=f(1+3)=f\left(1\right)+f\left(3\right)=7+21=28$ and so on

$\therefore f\left(n\right)=7+(n-1)7=7n$

Now, $\sum _{r=1}^{n}f\left(r\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+....+f\left(n\right)$

$=7+14+21+....+7n=7(1+2+3+....+n)$

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