You said that you want to practice summation problems! Didn't you?

$\large \displaystyle S = \lim _{n \rightarrow \infty} \sum_{m=0}^{n} \left [ 3 \left ( 1 + \frac{m}{n} \right ) ^2 \frac{1}{n} \right ]$

This turns out to be a Riemann Sum , whose limit is an integral. The conversion formula is:

$\large \displaystyle \lim _{n \rightarrow \infty} \sum_{m=0}^{n} f \left ( a + \frac{b-a}{n} \cdot m \right ) \frac{b-a}{n} = \int_a^b f(x) dx$

In our case, $a=0$ and $b=1$ , so:

$\large \displaystyle S = \int_0^1 3 (1 +x)^2 dx$

$\large \displaystyle S = (3x + 3x^2 + x^3) \Bigg |_0^1$

$\color{#3D99F6} \boxed{ \large \displaystyle S = 7}$

$\displaystyle \lim_{n \to \infty} \sum^{n}_{m=1} 3 \left(1+\frac{m} {n}\right)^{2}\dfrac{1}{n} \\ \displaystyle \lim_{n \to \infty} \frac{3}{n} \sum^{n}_{m=1} \left(1+\frac{m}{n}\right)^{2} \\ \displaystyle \lim_{n \to \infty} \frac{3}{n} \sum^{n}_{m=1} \left(1+\frac{2m}{n}+\frac{m^{2}}{n^{2}}\right) \\ \displaystyle \lim_{n \to \infty} \frac{1}{3} \left( \sum^{n}_{m=1} 1+\frac{2}{n} \left(\sum^{n}_{m=1}m\right)+\frac{1}{n^{2}} \left(\sum^{n}_{m=1}m^{2}\right)\right) \\ \displaystyle \lim_{n \to \infty} \frac{1}{3} \left(n+\frac{2}{n} \left(\frac{n(n+1)}{2}\right)+\frac{1}{n^{2}} \left(\frac{n(n+1)(2n+1)}{6}\right)\right) \\ \displaystyle \lim_{n \to \infty}\frac{3}{n} \left(2n+1+\frac{2n^{2}+3n+1}{6n}\right) \\ \displaystyle \lim_{n \to \infty} \left(6+\frac{3}{n}+\frac{6n^{2}+9n+3}{6n^{2}}\right) \\ \displaystyle \lim_{n \to \infty}\left(6+\frac{3}{n}+1+\frac{3}{2n}+\frac{1}{2n^{2}}\right) \\ \Rightarrow 6+0+1+0+0=\boxed{7}$