Question

If $f(x+y+z)=f\left(x\right)+f\left(y\right)+f\left(z\right)$ with $f\left(1\right)=1$ and $f\left(2\right)=2$ and $x,y,zϵR,$ then evaluate $\underset{n\to \infty }{lim}\frac{3\sum _{r=1}^n\left(4r\right)f\left(3r\right)}{n^3}$ is equal to

Answer 1

Let $x=y=z=1$
$\Rightarrow f\left(3\right)=3\times f\left(1\right)=3$
Let $y=z=1$
$\Rightarrow f(x+2)=f\left(x\right)+2$.
Hence,for integral values of $x$, $f\left(x\right)=x$
Let $L=\underset{n\to \infty }{lim}\frac{3\sum _{r=1}^n\left(4r\right)f\left(3r\right)}{n^3}$
$L=\underset{n\to \infty }{lim}\frac{3\sum _{r=1}^n\left(4r\right)\left(3r\right)}{n^3}$
$L=\underset{n\to \infty }{lim}36\frac{\left(n\right)(n+1)(2n+1)}{6n^3}$
$L=\frac{36\times 2}{6}$
$L=12$