Question

Let $N$ be the set of natural numbers. Suppose $f:N\to N$ is a function satisfying the following conditions
$\left(a\right)f(m+n)=f\left(m\right)+f\left(n\right)\left(b\right)f\left(2\right)=2$
The value of $\frac{1}{7}\sum _{k=1}^{20}f\left(k\right)$ is

Answer 1

$f:N\to Nf(m+n)=f\left(m\right)+f\left(n\right)$
Putting $m=0,n=0$, we get
$f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0$
$m=n=1$, we get
$f\left(2\right)=2f\left(1\right)\Rightarrow f\left(1\right)=1$
$m=2,n=1$, we get
$f\left(3\right)=f\left(1\right)+f\left(2\right)=3$
Similarly,
$f\left(4\right)=4$
Therefore,
$f\left(x\right)=x$
Now,
$\frac{1}{7}\sum _{k=1}^{20}f\left(k\right)=\frac{1}{7}\sum _{k=1}^{20}k=\frac{20\times 21}{7\times 2}=30$