How to solve $\lim_{n \to \infty} \overset{n}\sum_{k=1} 3(1+\frac{2k}{n})\frac{2}{n}$?

$$\sum_{k=1}^n 3\left(1+\frac{2k}{n}\right)\frac{2}{n}=\frac6n\left(\sum_{k=1}^n{1}+\frac2n\sum_{k=1}^{n}{k}\right)$$ $$=\frac6n\left(n+\frac2n\times\frac{n(n+1)}{2}\right)=\frac6n(n+n+1)$$ $$=6\left(\frac{2n+1}{n}\right)=6\left(\frac{2n}{n}+\frac1n\right)=6\left(2+\frac1n\right)$$

This is a Riemann sum of the function $x \mapsto 6\left(1+2x\right)$ over the interval $[0,1]$.

As this map is continuous and therefore Riemann integrable, this limit exists and is equal to $$\int_0^1 6(1+2x) \ dx =12$$