Question

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

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

Answer 1

Answer: (C)

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

$f\left(x\right)=ax$ where 'a' is constant.
$f(x+y)=a(x+y)$
$=ax+ay=f\left(x\right)+f\left(y\right)$
Now
$f\left(1\right)=7$
Or
$a=7$
Hence $f\left(x\right)=7x$
${\sum }_{n=1}^{n}f\left(x\right)$
$=7(1+2+3+..n)$
$=7\left(\frac{n(n+1)}{2}\right)$.