Question

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

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

Answer 1

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

Given,

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

So, Now

$f(1+1)=f\left(1\right)+f\left(1\right)$ $[Since,f(a+b)=f(a)+f(b\left)\right]$

$i.e.f\left(2\right)=7+7$

$so,f\left(2\right)=2\times 7$

Similarly,

$f\left(3\right)=3\times 7$

$f\left(n\right)=n\times 7$

Now,

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

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