Question

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

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

Answer 1

Answer: (D)

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

Putting x=0, y=1, we have
$f(0+1)=f\left(0\right)+f\left(1\right)\Rightarrow f\left(0\right)=0$

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

$\therefore f\left(2\right)=f(1+1)=f\left(1\right)+f\left(1\right)=7.2$

$f\left(3\right)=f(2+1)=f\left(2\right)+f\left(1\right)=7.3$ etc.

We know, ${\sum }_{i=1}^{n}k=1+2+3+\dots +n=\frac{n(n+1)}{2}$ (sum of first n natural numbers)

$\therefore {\sum }_{i=1}^{n}f\left(r\right)=7(1+2+3+.......+n)=\frac{7n(n+1)}{2}$