How to solve $\lim_{n \to \infty} \overset{n}\sum_{k=1} 3(1+\frac{2k}{n})\frac{2}{n}$? I need to solve $\lim_{n \to \infty} \overset{n}\sum_{k=1} 3(1+\frac{2k}{n})\frac{2}{n}$
I have worked to this by trying to calculate $\int^3_{1}3xdx$, but I am not sure how to get rid of the $k$ and find the closed form of the equation.
$$\Delta x= \frac{b-a}{n}=\frac{2}{n}$$
$$x_{k}=a+\Delta x_{k}= 1 + \frac{2k}{N}$$
$$\lim_{n \to \infty} \overset{n}\sum_{k=1}f(x_{k})\Delta x=\lim_{n \to \infty} \overset{n}\sum_{k=1} 3(1+\frac{2k}{n})\frac{2}{n}$$
 A: $$\sum_{k=1}^n 3\left(1+\frac{2k}{n}\right)\frac{2}{n}=\frac6n\left(\sum_{k=1}^n{1}+\frac2n\sum_{k=1}^{n}{k}\right)$$
$$=\frac6n\left(n+\frac2n\times\frac{n(n+1)}{2}\right)=\frac6n(n+n+1)$$
$$=6\left(\frac{2n+1}{n}\right)=6\left(\frac{2n}{n}+\frac1n\right)=6\left(2+\frac1n\right)$$
A: 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
$$\int_0^1 6(1+2x) \ dx =12$$
A: If you are trying to turn the given sum into a Riemann sum then into an integral, remember that
$$\int_a^b f(x)dx=\lim_{n\to\infty}\sum_{k=1}^{n}\frac{b-a}nf\left(a+\frac{k}{n}(b-a)\right)\\
=\lim_{n\to\infty}\frac{b-a}n\sum_{k=1}^{n}f\left(a+\frac{k}{n}(b-a)\right)$$
so we see that 
$$\frac{b-a}{n}=\frac2n\Rightarrow b-a=2$$
and 
$$f\left(a+\frac{k}{n}(b-a)\right)=f\left(a+\frac{2k}{n}\right)=3\left(1+\frac{2k}{n}\right)\Rightarrow f(x)=3x,\text{ and }a=1$$
Hence $b=3$. Therefore 
$$\lim_{n\to\infty}\frac2n\sum_{k=1}^{n}3\left(1+\frac{2k}{n}\right)=\int_1^3 3xdx=3\int_1^3xdx=\frac32(3^2-1)=\frac32\cdot8=12$$
