Question

If $f\left(x\right)$ satisfies the relation $f(x+y)=f\left(x\right)+f\left(y\right)$ for all $x,y\in R$, and $f\left(1\right)=5$, then

• A. $f\left(x\right)$ is an odd function
• B. $f\left(x\right)$ is an even function
• C. $\sum _{r=1}^{m}f\left(r\right)=5^{m+1}{C}_2$
• D. $\sum _{r=1}^{m}f\left(r\right)=\frac{5m(m+2)}{3}$

Answer 1

Answer: (A), (C)

$f\left(x\right)$ is an odd function
$\sum _{r=1}^{m}f\left(r\right)=5^{m+1}{C}_2$

Let $x=y=0$, then the given equation is :
$f\left(0\right)=2\times f\left(0\right)\Rightarrow f\left(0\right)=0$
Now for $y=-x$, we have
$0=f\left(0\right)=f\left(x\right)+f(-x)$
$\therefore f\left(x\right)=-f(-x)$
Hence the function is an odd function.
Now, we will prove that $f\left(k\right)=5k$ for any natural number $k$
Let $x=y=1$, then $f\left(2\right)=f\left(1\right)+f\left(1\right)=2\times 5$
Let $x=2,y=1$, then $f\left(3\right)=f\left(2\right)+f\left(1\right)=3\times 5$
In general,
Let $x=k-1,y=1$, then $f\left(k\right)=f(k-1)+f\left(1\right)=k\times 5$
Hence,
$\sum _{r=1}^{m}f\left(r\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)+\cdots +f\left(m\right)$
$=5+10+15+\cdots 5m=5\left(1+2+3+\cdots +m\right)=\frac{5m\left(m+1\right)}{2}=5\left({}^{m+1}{C}_2\right)$